Additional Information and Full Hypothesis Test Examples, Hypothesis Testing with One Sample | Statistics

Learning objectives

In a hypothesis test problem, you may see words such as "the level of significance is 1 percent". The "1 percent" is the preconceived or preset α.
The statistician setting up the hypothesis test selects the value of α to use before collecting the sample data.
If no level of significance is given, a common standard to use is α = 0.05.
When you calculate the p-value and draw the picture, the p-value is the area in the left tail, the right tail, or split evenly between the two tails. For this reason, we call the hypothesis test left, right, or two tailed.
The alternative hypothesis, $H_{a}$ , tells you if the test is left, right, or two-tailed. It is the key to conducting the appropriate test.
H_a never has a symbol that contains an equal sign.
Thinking about the meaning of the p-value: A data analyst should have more confidence that he made the correct decision to reject the null hypothesis with a smaller p-value (for example, 0.001 as opposed to 0.04) even if using the 0.05 level for alpha. Similarly, for a large p-value such as 0.4, as opposed to a p-value of 0.056 (alpha = 0.05 is less than either number), a data analyst should have more confidence that she made the correct decision in not rejecting the null hypothesis. This makes the data analyst use judgment rather than mindlessly applying rules.

The following examples illustrate a left-, right-, and two-tailed test.

H₀: μ = 5 H_a: μ < 5

Test of a single population mean. H_a tells you the test is left-tailed. The picture of the p-value is as follows:

View original figure (external; image not reproduced in this text edition)

Original figure description: Normal distribution curve of a single population mean with a value of 5 on the x-axis and the p-value points to the area on the left tail of the curve.

H₀: p ≤ 0.2 H_a: p > 0.2

This is a test of a single population proportion. H_a tells you the test is right-tailed. The picture of the p-value is as follows:

View original figure (external; image not reproduced in this text edition)

Original figure description: Normal distribution curve of a single population proportion with the value of 0.2 on the x-axis. The p-value points to the area on the right tail of the curve.

H₀: p = 50 H_a: p ≠ 50

This is a test of a single population mean. H_a tells you the test is two-tailed. The picture of the p-value is as follows.

View original figure (external; image not reproduced in this text edition)

Original figure description: Normal distribution curve of a single population mean with a value of 50 on the x-axis. The p-value formulas, 1/2(p-value), for a two-tailed test is shown for the areas on the left and right tails of the curve.

Full Hypothesis Test Examples

Jeffrey, as an eight-year-old, established a mean time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds. His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-yard freestyle swims. For the 15 swims, Jeffrey's mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds. Conduct a hypothesis test using a preset α = 0.05. Assume that the swim times for the 25-yard freestyle are normal.

Set up the hypothesis test:

Since the problem is about a mean, this is a test of a single population mean.

H₀: μ = 16.43 H_a: μ < 16.43

For Jeffrey to swim faster, his time will be less than 16.43 seconds. The "<" tells you this is left-tailed.

Determine the distribution needed:

Random variable: $X$ = the mean time to swim the 25-yard freestyle.

Distribution for the test: $X$ is normal (population standard deviation is known: σ = 0.8)

with mean $μ = 16.43$ and standard error of $\frac{0.8}{\sqrt{15}}$ ;

μ = 16.43 comes from H₀ and not the data. σ = 0.8, and n = 15.

Using a table or a calculator, we can calculate the p-value as the area to the left of 16 under the normal curve:

p-value = P( $\bar{x}$ < 16) = 0.0187 where the sample mean in the problem is given as 16.

p-value = 0.0187. The p-value is the area to the left of the sample mean given as 16.

Graph:

View original figure (external; image not reproduced in this text edition)

Original figure description: Normal distribution curve for the average time to swim the 25-yard freestyle with values 16, as the sample mean, and 16.43 on the x-axis. A vertical upward line extends from 16 on the x-axis to the curve. An arrow points to the left tail of the curve.

μ = 16.43 comes from H₀. Our assumption is μ = 16.43.

Interpretation of the p-value: If H₀ is true, there is a 0.0187 probability (1.87 percent), that Jeffrey's mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87 percent chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event.

Compare α and the p-value:

α = 0.05 p-value = 0.0187 α > p-value

Make a decision: Since α > p-value, reject H₀.

An alternative approach is to find the z-test corresponding to the sample mean $\bar{x} = 16.$ This is $z -test = \frac{\bar{x} - μ_{X}}{\frac{σ_{X}}{\sqrt{n}}} = \frac{16 - 16.43}{\frac{0.8}{\sqrt{15}}} = - 2.081729$ .

The critical z-value = –1.645 for this test has probability 0.05 to its left tail, according to the Normal Table (see Appendices). Because the z-test is to the left of the critical z-value, we reject the null hypothesis.

This means that you reject μ = 16.43. In other words, you do not think Jeffrey swims the 25-yard freestyle in 16.43 seconds but instead that he swims faster with the new goggles.

Conclusion: At the 5 percent significance level, we conclude that Jeffrey swims faster using the new goggles. The sample data show there is sufficient evidence that Jeffrey's mean time to swim the 25-yard freestyle is less than 16.43 seconds.

The p-value can easily be calculated.

When the calculator does a z-Test, the z-Test function finds the p-value by doing a normal probability calculation:

$P (\bar{x} < 16) =$ 2nd DISTR normcdf $(- 10^99, 16, 16.43, 0.8 / \sqrt{15}) .$

The Type I and Type II errors for this problem are as follows:

The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.)

The Type II error is that there is not evidence to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yard freestyle, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.)

Historical Note (fs-idp41487520)

The traditional way to compare the two probabilities, α and the p-value, is to compare the critical value (z-score from α) to the test statistic (z-score from data). The calculated test statistic for the p-value is –2.08. (From the central limit theorem, the test statistic formula is $z = \frac{\bar{x} - μ_{X}}{(\frac{σ_{X}}{\sqrt{n}})}$ . For this problem, $\bar{x}$ = 16, μ_X = 16.43 from the null hypothesis, σ_X = 0.8, and n = 15.) You can find the critical value for α = 0.05 in the normal table (see Appendix H: Tables). The z-score for an area to the left equal to 0.05 is midway between –1.65 and –1.64 (0.05 is midway between 0.0505 and 0.0495). The z-score is –1.645. Since –1.645 > –2.08 (which demonstrates that α > p-value), reject H₀. Traditionally, the decision to reject or not reject was done in this way. Today, comparing the two probabilities α and the p-value is very common. For this problem, the p-value, 0.0187, is considerably smaller than α, 0.05. You can be confident about your decision to reject. The graph shows α, the p-value, and the test statistic and the critical value.

View original figure (external; image not reproduced in this text edition)

Original figure description: Distribution curve comparing the α to the p-value. Values of -2.15 and -1.645 are on the x-axis. Vertical upward lines extend from both of these values to the curve. The p-value is equal to 0.0158 and points to the area to the left of -2.15. α is equal to 0.05 and points to the area between the values of -2.15 and -1.645.

Try It

The mean throwing distance of a football by Marco, a high school freshman quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco’s mean distance was 45 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Conduct a hypothesis test using a preset α = 0.05. Assume the throw distances for footballs are normal.

First, determine what type of test this is, set up the hypothesis test, find the p-value, sketch the graph, and state your conclusion.

Since the problem is about a mean, this is a test of a single population mean.

H₀ : μ = 40

H_a : μ > 40

p = 0.0062

View original figure (external; image not reproduced in this text edition)

Original figure description: A graph showing a bell shaped curve of normal distribution with one vertical line to the right of center. The vertical lines are not labeled. The area to the right of the vertical line is shaded purple. A note above the shaded areas says p value. The horizontal axis is labeled as the mean. There are two tick marks on the horizontal axis. One tick mark is at the center of the bell shaped curve labeled 40 and the other is at the vertical line labeled 45.

Because p < α, we reject the null hypothesis. There is sufficient evidence to suggest that the change in grip improved Marco’s throwing distance.

A college football coach records the mean weight that his players can bench press as 275 pounds, with a standard deviation of 55 pounds. Three of his players thought that the mean weight was more than that amount. They asked 30 of their teammates for their estimated maximum lift on the bench press exercise. The data ranged from 205 pounds to 385 pounds. The actual different weights were (frequencies are in parentheses)

205(3)
215(3)
225(1)
241(2)
252(2)
265(2)
275(2)
313(2)
316(5)
338(2)
341(1)
345(2)
368(2)
385(1)

Conduct a hypothesis test using a 2.5 percent level of significance to determine if the bench press mean is more than 275 pounds.

Set up the hypothesis test:

Since the problem is about a mean weight, this is a test of a single population mean.

H₀: μ = 275 H_a: μ > 275 This is a right-tailed test.

Calculating the distribution needed:

Random variable: $\bar{X}$ = the mean weight, in pounds, lifted by the football players.

Distribution for the test: It is normal because σ is known.

$\bar{X} ~ N (275, \frac{55}{\sqrt{30}})$

$\bar{x} = 286.2$ pounds (from the data).

σ = 55 pounds. Always use σ if you know it. We assume μ = 275 pounds unless our data shows us otherwise.

First, we compute the sample mean:

\bar{x} = \frac{205 + 205 + 205 + 215 + \dots + 385}{30} = 286.2.

Next, we compute the z-test:

z -test = \frac{286.2 - 275}{\frac{55}{\sqrt{30}}} = 1.115362

Finally, the p-value is the probability to the right tail of the z-test, which we can compute from the table of z-scores as 0.5 - 0.36650 = 0.1335.

$p -value = P (\bar{x} > 286.2) = 0.1323$

Interpretation of the p-value: If H₀ is true, then there is a 0.1331 probability, 13.23 percent, that the football players can lift a mean weight of 286.2 pounds or more. Because a 13.23 percent chance is large enough, a mean weight lift of 286.2 pounds or more is not a rare event.

View original figure (external; image not reproduced in this text edition)

Original figure description: Normal distribution curve of the average weight lifted by football players with values of 275 and 286.2 on the x-axis. A vertical upward line extends from 286.2 to the curve. The p-value points to the area to the right of 286.2.

Compare α and the p-value:

α = 0 .025

p -value = 0 .1323

Make a decision: Since α < p-value, do not reject H₀.

Conclusion: At the 2.5 percent level of significance, from the sample data, there is not sufficient evidence to conclude that the true mean weight lifted is more than 275 pounds.

The p-value can easily be calculated.

Put the data and frequencies into lists. Press STAT and arrow over to TESTS. Press 1:z-Test. Arrow over to Data and press ENTER. Arrow down and enter 275 for μ₀, 55 for σ, the name of the list where you put the data, and the name of the list where you put the frequencies. Arrow down to μ: and arrow over to > μ₀. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value (p = 0.1331, a little different from the previous calculation—in it we used the sample mean rounded to one decimal place instead of the data), but also the test statistic (z-score) for the sample mean, the sample mean, and the sample standard deviation. μ > 275 is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with z = 1.112 (test statistic) and p = 0.1331 (p-value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off.

Statistics students believe that the mean score on the first statistics test is 65. A statistics instructor thinks the mean score is higher than 65. He samples 10 statistics students and obtains the scores

. He performs a hypothesis test using a 5 percent level of significance. The data are assumed to be from a normal distribution.

Set up the hypothesis test:

A 5 percent level of significance means that α = 0.05. This is a test of a single population mean.

H₀: μ = 65 H_a: μ > 65

Since the instructor thinks the average score is higher, use a ">". The ">" means the test is right-tailed.

Determine the distribution needed:

Random variable: $\bar{X}$ = average score on the first statistics test.

Distribution for the test: If you read the problem carefully, you will notice that there is no population standard deviation given. You are only given n = 10 sample data values. Notice also that the data come from a normal distribution. This means that the distribution for the test is a Student's t-distribution.

Use t-distribution. Therefore, the distribution for the test is t with nine degrees of freedom.

Calculate the p-value using the Student's t-distribution:

First, we compute the sample mean as

\bar{x} = \frac{65 + 65 + \dots + 71}{10} = 67.

Next, we compute the t-test as

t -test = \frac{\bar{x} - μ_{X}}{\frac{s_{X}}{\sqrt{n}}} = \frac{67 - 65}{\frac{3 .12}{\sqrt{10}}} \approx 1 .98 .

The p-value is the probability to the right tail of 1.98 in a t-distribution with nine degrees of freedom.

p-value = P( $\bar{x}$ > 67) = 0.0396 where the sample mean and sample standard deviation are calculated as 67 and 3.1972 from the data.

Interpretation of the p-value: If the null hypothesis is true, then there is a 0.0396 probability (3.96 percent) that the sample mean is 65 or more.

View original figure (external; image not reproduced in this text edition)

Original figure description: Normal distribution curve of average scores on the first statistic tests with 65 and 67 values on the x-axis. A vertical upward line extends from 67 to the curve. The p-value points to the area to the right of 67.

Compare α and the p-value:

Since α = 0.05 and p-value = 0.0396, α > p-value.

Make a decision: Since α > p-value, reject H₀.

Alternatively, according to a Student's t-distribution table (see Appendices), the critical t-value is 1.833. Since the t-test (1.98) is to the right of the critical t-value 1.833, we reject the null hypothesis.

This decision means we reject μ = 65. In other words, we believe the average test score is more than 65.

Conclusion: At a 5 percent level of significance, the sample data show sufficient evidence that the mean (average) test score is more than 65, just as the math instructor thinks.

The p-value can easily be calculated.

Put the data into a list. Press STAT and arrow over to TESTS. Press 2:T-Test. Arrow over to Data and press ENTER. Arrow down and enter 65 for μ₀, the name of the list where you put the data, and 1 for Freq:. Arrow down to μ: and arrow over to > μ₀. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value (p = 0.0396) but it also calculates the test statistic (t-score) for the sample mean, the sample mean, and the sample standard deviation. μ > 65 is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with t = 1.9781 (test statistic) and p = 0.0396 (p-value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off.

Try It

It is believed that a stock price for a particular company will grow at a rate of $5 per week with a standard deviation of $1. An investor believes the stock won’t grow as quickly. The changes in stock price are recorded for 10 weeks and are as follows: $4, $3, $2, $3, $1, $7, $2, $1, $1, $2. Perform a hypothesis test using a 5 percent level of significance. State the null and alternative hypotheses, find the p-value, state your conclusion, and identify the Type I and Type II errors.

H₀: μ = 5

H_a: μ < 5

p = 0.0082

Because p < α, we reject the null hypothesis. There is sufficient evidence to suggest that the stock price of the company grows at a rate less than $5 a week.

Type I Error: To conclude that the stock price is growing slower than $5 a week when, in fact, the stock price is growing at $5 a week (reject the null hypothesis when the null hypothesis is true).

Type II Error: To conclude that the stock price is growing at a rate of $5 a week when, in fact, the stock price is growing slower than $5 a week (do not reject the null hypothesis when the null hypothesis is false).

Joon believes that 50 percent of first-time brides in the United States are younger than their grooms. She performs a hypothesis test to determine if the percentage is the same or different from 50 percent. Joon samples 100 first-time brides and 53 reply that they are younger than their grooms. For the hypothesis test, she uses a 1 percent level of significance.

Set up the hypothesis test:

The 1 percent level of significance means that α = 0.01. This is a test of a single population proportion.

H₀: p = 0.50 H_a: p ≠ 0.50

The words is the same or different from tell you this is a two-tailed test.

Calculate the distribution needed:

Random variable: P′ = the percentage of first-time brides who are younger than their grooms.

Distribution for the test: The problem contains no mention of a mean. The information is given in terms of percentages. Use the distribution for P′, the estimated proportion.

P′ follows a normal distribution with mean value $μ = p,$ and standard error $σ = \sqrt{\frac{p \cdot q}{n} .}$

In our example, p = q = 0.5, and n = 100,
where p = 0.50, q = 1 – p = 0.50, and n = 100.

Calculate the p-value using the normal distribution for proportions:

First, we compute the sample proportion as $\hat{p} = \frac{53}{100} = 0.53.$

Next, the z-test is given by

z -test = \frac{\hat{p} - p}{\sqrt{\frac{p \cdot q}{n}}} = \frac{0.53 - 0.50}{\sqrt{\frac{0.50 \times 0.50}{100}}} = 0.6 .

Since the z-test is positive, we compute the area to the right tail of 0.6 in a normal distribution, $P (Z > 0.6) = 0 .2742531 .$ Finally, because this is a two-sided test of significance, we multiply this probability times two to account for the left tail, and obtain $p -value = 2 \times 0.2742531 = 0.5485062$

where x = 53, p′ = $\frac{x}{n} = \frac{53}{100}$ = 0.53.

Interpretation of the p-value: If the null hypothesis is true, there is 0.5485 probability, (54.85 percent) that the sample (estimated) proportion p′ is 0.53 or more OR 0.47 or less (see the graph in hyptest11_ex5).

View original figure (external; image not reproduced in this text edition)

Original figure description: Normal distribution curve of the percent of first time brides who are younger than the groom with values of 0.47, 0.50, and 0.53 on the x-axis. Vertical upward lines extend from 0.47 and 0.53 to the curve. 1/2(p-values) are calculated for the areas on outsides of 0.47 and 0.53.

μ = p = 0.50 comes from H₀, the null hypothesis.

p′ = 0.53. Since the curve is symmetrical and the test is two-tailed, the p′ for the left tail is equal to 0.50 – 0.03 = 0.47 where μ = p = 0.50. (0.03 is the difference between 0.53 and 0.50.)

Compare α and the p-value:

Since α = 0.01 and p-value = 0.5485, α < p-value.

Make a decision: Since α < p-value, you cannot reject H₀.

Conclusion: At the 1 percent level of significance, the sample data do not show sufficient evidence that the percentage of first-time brides who are younger than their grooms is different from 50 percent.

The p-value can easily be calculated.

The Type I and Type II errors are as follows:

The Type I error is to conclude that the proportion of first-time brides who are younger than their grooms is different from 50 percent when, in fact, the proportion is actually 50 percent. Reject the null hypothesis when the null hypothesis is true.

The Type II error is there is not enough evidence to conclude that the proportion of first-time brides who are younger than their grooms differs from 50 percent when, in fact, the proportion does differ from 50 percent. Do not reject the null hypothesis when the null hypothesis is false.

Try It

A teacher believes that 85 percent of students in the class will want to go on a field trip to the local zoo. She performs a hypothesis test to determine if the percentage is the same or different from 85 percent. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1 percent level of significance.

First, determine what type of test this is, set up the hypothesis test, find the p-value, sketch the graph, and state your conclusion.

Since the problem is about percentages, this is a test of single population proportions.

H₀ : p = 0.85

H_a: p ≠ 0.85

p = 0.7554

View original figure (external; image not reproduced in this text edition)

Original figure description: A graph showing a bell shaped curve of normal distribution with vertical lines to the left and right of center. The areas on the outside legs of the vertical lines are shaded purple. A note above the shaded areas say one half p value on both sides. The horizontal axis is labeled as the p prime. There is one tick mark at the center of the horizontal axis labeled 0.85 and a tick mark at the bottom of the left vertical line labeled 0.78.

Because p > α, we fail to reject the null hypothesis. There is not sufficient evidence to suggest that the proportion of students that want to go to the zoo is not 85 percent.

Suppose a consumer group suspects that the proportion of households that have three cell phones is 30 percent. A cell phone company has reason to believe that the proportion is not 30 percent. Before the cell phone company starts a big advertising campaign, it conducts a hypothesis test. The company's marketing people survey 150 households with the result that 43 of the households have three cell phones.

The value that helps determine the p-value is p′. Calculate p′.
What is a success for this problem?
What is the level of significance?
Draw the graph for this problem. Draw the horizontal axis. Label and shade appropriately.
Calculate the p-value.
Make a decision. _____________(Reject/Do not reject) H₀ because____________.

Set up the hypothesis test:

H₀: p = 0.30 H_a: p ≠ 0.30

Determine the distribution needed:

The random variable is P′ = proportion of households that have three cell phones.

The distribution for the hypothesis test is $P' ~ N (0.30, \sqrt{\frac{(0.30) \cdot (0.70)}{150}}) .$

p′ = $\frac{x}{n}$ where x is the number of successes and n is the total number in the sample.
$x = 43, n = 150$ $p ′ = \frac{43}{150 .}$
A success is having three cell phones in a household.
The level of significance is the preset α. Since α is not given, assume that α = 0.05.
First we compute the sample proportion $\hat{p} = \frac{43}{150} = 0 .287 .$

Next, the z-test is given by
$z -test = \frac{\hat{p} - p}{\sqrt{\frac{p \cdot q}{n}}} = \frac{0.287 - 0.30}{\sqrt{\frac{0.30 \times 0.70}{150}}} \approx –0.36 .$
Since the z-test is negative, we compute the area to the left tail of –0.36 in a normal distribution, $P (Z < - 0.36) \approx 0 .3607902 .$ Finally, because this is a two-sided test of significance, we multiply this probability times two to account for the right tail, and obtain $p -value = 2 \times 0.3607902 = 0.7215804 .$
Assuming that α = 0.05, α < p-value. The decision is do not reject H₀ because there is not sufficient evidence to conclude that the proportion of households that have three cell phones is not 30 percent.

Try It

Marketers believe that 92 percent of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. Two hundred American adults are surveyed, of which 174 report having cell phones. Use a 5 percent level of significance. State the null and alternative hypotheses, find the p-value, state your conclusion, and identify the Type I and Type II errors.

H₀: p = 0.92

H_a: p < 0.92

p-value = 0.0046

Because p < 0.05, we reject the null hypothesis. There is sufficient evidence to conclude that fewer than 92 percent of American adults own cell phones.

Type I Error: To conclude that fewer than 92 percent of American adults own cell phones when, in fact, 92 percent of American adults do own cell phones (reject the null hypothesis when the null hypothesis is true).

Type II Error: To conclude that 92 percent of American adults own cell phones when, in fact, fewer than 92 percent of American adults own cell phones (do not reject the null hypothesis when the null hypothesis is false).

The next example is a poem written by a statistics student named Nicole Hart. The solution to the problem follows the poem. Notice that the hypothesis test is for a single population proportion. This means that the null and alternate hypotheses use the parameter p. The distribution for the test is normal. The estimated proportion p′ is the proportion of fleas killed to the total fleas found on Fido. This is sample information. The problem gives a preconceived α = 0.01, for comparison, and a 95 percent confidence interval computation. The poem is clever and humorous, so please enjoy it!

My dog has so many fleas,
They do not come off with ease.
As for shampoo, I have tried many types
Even one called Bubble Hype,
Which only killed 25 percent of the fleas,
Unfortunately I was not pleased.
I've used all kinds of soap,
Until I had given up hope
Until one day I saw
An ad that put me in awe.
A shampoo used for dogs
Called GOOD ENOUGH to Clean a Hog
Guaranteed to kill more fleas.
I gave Fido a bath
And after doing the math
His number of fleas
Started dropping by 3's!
Before his shampoo
I counted 42.
At the end of his bath,
I redid the math
And the new shampoo had killed 17 fleas.
So now I was pleased.
Now it is time for you to have some fun
With the level of significance being .01,
You must help me figure out
Use the new shampoo or go without?

Set up the hypothesis test:

H₀: p ≤ 0.25 H_a: p > 0.25

Determine the distribution needed:

In words, clearly state what your random variable $\bar{X}$ or P′ represents.

P′ = The proportion of fleas that are killed by the new shampoo

State the distribution to use for the test.

Normal:

N (0.25, \sqrt{\frac{(0.25) (1 - 0.25)}{42}})

The z-test is given by

z -test = \frac{\bar{p} - p}{\sqrt{\frac{p \cdot q}{n}}} = \frac{0.4048 - 0.25}{\sqrt{42}} \approx 2.316834 .

Because this is a hypothesis test one-sided to the right, we compute the p-value as the area to the right tail of the z-test in a standard normal distribution, $P (Z > 3.32) \approx 0 .0103 .$

In one to two complete sentences, explain what the p-value means for this problem.

If the null hypothesis is true (the proportion is 0.25), then there is a 0.0103 probability that the sample (estimated) proportion is 0.4048 $(\frac{17}{42})$ or more.

Use the previous information to sketch a picture of this situation. Clearly label and scale the horizontal axis and shade the region(s) corresponding to the p-value.

View original figure (external; image not reproduced in this text edition)

Original figure description: Normal distribution graph of the proportion of fleas killed by the new shampoo with values of 0.25 and 0.4048 on the x-axis. A vertical upward line extends from 0.4048 to the curve and the area to the left of this is shaded in. The test statistic of the sample proportion is listed.

Compare α and the p-value:

Indicate the correct decision (reject or do not reject the null hypothesis) and the reason for it, and write an appropriate conclusion, using complete sentences.

Alpha	Decision	Reason for Decision
0.01	Do not reject $H_{0}$	α < p-value

Conclusion: At the 1 percent level of significance, the sample data do not show sufficient evidence that the percentage of fleas that are killed by the new shampoo is more than 25 percent.

Construct a 95 percent confidence interval for the true mean or proportion. Include a sketch of the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval.

View original figure (external; image not reproduced in this text edition)

Original figure description: Normal distribution graph of the proportion of fleas killed by the new shampoo with values of 0.26, 17/42, and 0.55 on the x-axis. A vertical upward line extends from 0.26 and 0.55. The area between these two points is equal to 0.95.

Confidence Interval: (0.26, 0.55). We are 95 percent confident that the true population proportion p of fleas that are killed by the new shampoo is between 26 percent and 55 percent.

The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass:

1.11, 1.07, 1.11, 1.07, 1.12, 1.08, 0.98, 0.98, 1.02, 0.95, 0.95

Is there convincing evidence that the average conductivity of this type of glass is greater than one? Use a significance level of 0.05. Assume the population is normal.

Let’s follow a four-step process to answer this statistical question.

State the question: We need to determine if, at a 0.05 significance level, the average conductivity of the selected glass is greater than one. Our hypotheses will be as follows:
1. H₀: μ ≤ 1
2. H_a: μ > 1
Plan: We are testing a sample mean without a known population standard deviation. Therefore, we need to use a Student's t-distribution. Assume the underlying population is normal.
Do the calculations: We will input the sample data into the TI-83 as follows.

View original figure (external; image not reproduced in this text edition)

Original figure description: The display of a T I 83 calculator is shown with three columns. The first column is L1 with the following values 1.08, .98, .98, 1.02, .95, .95. The second column is labeled L2 with no values listed. The third column is labeled L3 with no values. There is a 1 on the top row on the far right side of the screen.

View original figure (external; image not reproduced in this text edition)

Original figure description: The display of a T I 83 calculator is shown. The display is titled T Test. The display shows 6 rows. The first row reads Inpt colon data stats and data is highlighted. The second row reads mu not 1. The third row reads List L1. The fourth row reads Freq 1. The fifth row reads mu with the following options does not equal mu not, less than mu not, greater than mu not greater than mu not is highlighted. The sixth row reads Calculate Draw and neither option is highlighted.

View original figure (external; image not reproduced in this text edition)

Original figure description: The display of a T I 83 calculator is shown. The display is titled T Test. The display shows 6 rows. The first row reads mu greater than 1. The second row reads t equals 2.01377743. The third row reads p equals .035860646. The fourth row reads x bar equals 1.04. The fifth row reads S sub x equals .0658786764. The sixth row reads n equals 11.

View original figure (external; image not reproduced in this text edition)

Original figure description: The display of a T I 83 calculator is shown. The display is a bell shaped curve of normal distribution. The vertical axis has four tickmarks evenly spaced out and the top of the bell curve is at the fourth tick mark. The horizontal axis has two tick marks on each side of the bell curve. The area to the right of the second tick mark on the right portion of the curve is shaded. At the bottom of the display it is shown that t equals 2.0138 and that p equals .0359.
State the conclusions: Since the p-value (p = 0.036) is less than our alpha value, we will reject the null hypothesis. It is reasonable to state that the data support the claim that the average conductivity level is greater than one.

In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test the claim that cell phone users developed brain cancer at a greater rate than that for non-cell phone users. The rate of brain cancer for non-cell phone users is 0.0340 percent. Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.

We will follow the four-step process.

We need to conduct a hypothesis test on the claimed cancer rate. Our hypotheses will be as follows:
1. H₀: p ≤ 0.00034
2. H_a: p > 0.00034
If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes cancer-causing environments, we want to minimize the chances of incorrectly identifying causes of cancer.
We will be testing a sample proportion with x = 172 and n = 420,019. The sample is sufficiently large because we have np = 420,019(0.00034) = 142.8, nq = 420,019(0.99966) = 419,876.2, two independent outcomes, and a fixed probability of success p = 0.00034. Thus we will be able to generalize our results to the population.
The associated TI results are shown in the following figures.

View original figure (external; image not reproduced in this text edition)

Original figure description: The display of a T I 83 calculator is shown. The display is titled One Prop Z Test. The display shows 5 rows. The first row reads prop greater than 3.4 E negative 4. The second row reads z equals 2.443355592. The third row reads p equals .0072756984. The fourth row reads p bar equals 4.0950528 E negative 4. The fifth row reads n equals 420019.

View original figure (external; image not reproduced in this text edition)

Original figure description: The display of a T I 83 calculator is shown. The display is a bell shaped curve of normal distribution. The vertical axis has four tickmarks evenly spaced out and the top of the bell curve is at the fourth tick mark. The horizontal axis has two tick marks on each side of the bell curve. The area to the right of the second tick mark on the right portion of the curve is shaded. At the bottom of the display it is shown that t equals 2.4434 and that p equals .0073.
Since the p-value = 0.0073 is greater than our alpha value = 0.005, we cannot reject the null. Therefore, we conclude that there is not enough evidence to support the claim of higher brain cancer rates for the cell phone users.

Chapter Review

The hypothesis test itself has an established process. This can be summarized as follows:

Determine H₀ and H_a. Remember, they are contradictory.
Determine the random variable.
Determine the distribution for the test.
Draw a graph, calculate the test statistic, and use the test statistic to calculate the p-value. (A z-score and a t-score are examples of test statistics.)
Compare the preconceived α with the p-value, make a decision (reject or do not reject H₀), and write a clear conclusion using English sentences.

Notice that in performing the hypothesis test, you use α and not β. β is needed to help determine the sample size of the data that are used in calculating the p-value. Remember that the quantity 1 – β is called the Power of the Test. A high power is desirable. If the power is too low, statisticians typically increase the sample size while keeping α the same. If the power is low, the null hypothesis might not be rejected when it should be.

Assume H₀: μ = 9 and H_a: μ < 9. Is this a left-tailed, right-tailed, or two-tailed test?

This is a left-tailed test.

Assume H₀: μ ≤ 6 and H_a: μ > 6. Is this a left-tailed, right-tailed, or two-tailed test?

Assume H₀: p = 0.25 and H_a: p ≠ 0.25. Is this a left-tailed, right-tailed, or two-tailed test?

This is a two-tailed test.

Draw the general graph of a left-tailed test.

Draw the graph of a two-tailed test.

View original figure (external; image not reproduced in this text edition)

Original figure description: A graph showing a bell shaped curve of normal distribution with two vertical lines to the left and right of center. The areas on the outside legs of the vertical lines are shaded purple. A note above the shaded areas say one half p value on both sides. There is one tick mark in the center of the bell shaped curve labeled mu and the horizontal axis is labeled as the mean.

A bottle of water is labeled as containing 16 fluid ounces of water. You believe it is less than that. What type of test would you use?

Your friend claims that his mean golf score is 63. You want to show that it is higher than that. What type of test would you use?

a right-tailed test

A bathroom scale claims to be able to identify correctly any weight within a pound. You think that it cannot be that accurate. What type of test would you use?

You flip a coin and record whether it shows heads or tails. You know the probability of getting heads is 50 percent, but you think it is less for this particular coin. What type of test would you use?

a left-tailed test

If the alternative hypothesis has a not equals ( ≠ ) symbol, you know to use which type of test?

Assume the null hypothesis states that the mean is at least 18. Is this a left-tailed, right-tailed, or two-tailed test?

This is a left-tailed test.

Assume the null hypothesis states that the mean is at most 12. Is this a left-tailed, right-tailed, or two-tailed test?

Assume the null hypothesis states that the mean is equal to 88. The alternative hypothesis states that the mean is not equal to 88. Is this a left-tailed, right-tailed, or two-tailed test?

This is a two-tailed test.

Homework

For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E, Solution Sheets. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.

A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using alpha = 0.05, are the data highly inconsistent with the claim?

H₀: μ ≥ 50,000
H_a: μ < 50,000
Let $\bar{X}$ = the average lifespan of a brand of tires.
normal distribution
z = -2.315
p-value = 0.0103
Check student’s solution.
1. Alpha: 0.05
2. Decision: Reject the null hypothesis.
3. Reason for decision: The p-value is less than 0.05.
4. Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is less than 50,000 miles.
(43,537, 49,463)

In 2009, President Barack Obama announced a new national fuel economy and emissions policy for cars and light trucks. It stated that the combined fleet fuel economy for an auto manufacturer of cars and light trucks will have to average 35.5 mpg or better by 2016. From past studies on fuel economy, it is known that the standard deviation of a typical fleet is 7.6 mpg. An auto manufacturer selects a random sample of 55 cars and light trucks and finds the sample mean fuel economy to be 34.6 mpg with a standard deviation of 10.3 mpg. Can the manufacturer claim that their fleet meets the fuel economy standard in the 2016 policy at the 5 percent level?

H₀: μ ≥ 35.5
H_a: μ < 35.5
Let $\bar{x}$ = the average mpg for the sample of cars and trucks in the fleet
normal distribution
z = -0.648
p-value = 0.2578
Check student’s solution.
1. Alpha: 0.05
2. Decision: Do not reject the null hypothesis.
3. Reason for decision: The p-value is greater than 0.05.
4. Conclusion: There is sufficient evidence to support the claim that the manufacturer’s fleet meets the fuel economy standards in the 2016 policy.
(31.88 mpg, 37.32 mpg)

The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of 20¢. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. Twelve costs yield a mean cost of 95¢ with a standard deviation of 18¢. Do the data support the claim at the 1 percent level?

H₀: μ = $1.00
H_a: μ ≠ $1.00
Let $\bar{x}$ = the average cost of a daily newspaper.
normal distribution
z = –0.866
p-value = 0.3865
Check student’s solution.
1. Alpha: 0.01
2. Decision: Do not reject the null hypothesis.
3. Reason for decision: The p-value is greater than 0.01.
4. Conclusion: There is sufficient evidence to support the claim that the mean cost of daily papers is $1. The mean cost could be $1.
($0.84, $1.06)

An article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 49 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1 percent level?

The mean number of sick days an employee takes per year is believed to be about 10. Members of a personnel department do not believe this figure. They randomly survey eight employees. The number of sick days they took for the past year are as follows: 12; 4; 15; 3; 11; 8; 6; 8. Let x = the number of sick days they took for the past year. Should the personnel team believe that the mean number is 10?

H₀: μ = 10
H_a: μ ≠ 10
Let $\bar{X}$ = the mean number of sick days an employee takes per year.
Student’s t-distribution
t = –1.12
p-value = 0.300
Check student’s solution.
1. Alpha: 0.05
2. Decision: Do not reject the null hypothesis.
3. Reason for decision: The p-value is greater than 0.05.
4. Conclusion: At the 5 percent significance level, there is insufficient evidence to conclude that the mean number of sick days is not 10.
(4.9443, 11.806)

In 1955, Life Magazine reported that the 25-year-old mother of three worked, on average, an 80-hour week. Recently, many groups have been studying whether or not the women's movement has, in fact, resulted in an increase in the average work week for women (combining employment and at-home work). Suppose a study was done to determine if the mean work week has increased. Eighty-one women were surveyed with the following results. The sample mean was 83; the sample standard deviation was 10. Does it appear that the mean work week has increased for women at the 5 percent level?

Your statistics instructor claims that 60 percent of the students who take her Elementary Statistics class go through life feeling more enriched. For some reason that she can't quite figure out, most people don't believe her. You decide to check this out on your own. You randomly survey 64 of her past Elementary Statistics students and find that 34 feel more enriched as a result of her class. Now, what do you think?

H₀: p ≥ 0.6
H_a: p < 0.6
Let P′ = the proportion of students who feel more enriched as a result of taking elementary statistics.
normal for a single proportion
1.12
p-value = 0.1308
Check student’s solution.
1. Alpha: 0.05
2. Decision: Do not reject the null hypothesis.
3. Reason for decision: The p-value is greater than 0.05.
4. Conclusion: There is insufficient evidence to conclude that less than 60 percent of her students feel more enriched.
Confidence interval: (0.409, 0.654)
The “plus-4s” confidence interval is (0.411, 0.648)

A Nissan Motor Corporation advertisement read, “The average man’s I.Q. is 107. The average brown trout’s I.Q. is 4. So why can’t man catch brown trout?” Suppose you believe that the brown trout’s mean I.Q. is greater than four. You catch 12 brown trout. A fish psychologist determines the I.Q.s as follows: 5, 4, 7, 3, 6, 4, 5, 3, 6, 3, 8, 5. Conduct a hypothesis test of your belief.

Refer to element-410. Conduct a hypothesis test to see if your decision and conclusion would change if your belief were that the brown trout’s mean I.Q. is not four.

H₀: μ = 4
H_a: μ ≠ 4
Let $\bar{X}$ the average I.Q. of a set of brown trout.
two-tailed Student's t-test
t = 1.95
p-value = 0.076
Check student’s solution.
1. Alpha: 0.05
2. Decision: Reject the null hypothesis.
3. Reason for decision: The p-value is greater than 0.05
4. Conclusion: There is insufficient evidence to conclude that the average IQ of brown trout is not four.
(3.8865, 5.9468)

According to an article in Newsweek, the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100: 114 (46.7 percent girls). Suppose you don’t believe the reported figures of the percentage of girls born in China. You conduct a study. In this study, you count the number of girls and boys born in 150 randomly chosen recent births. There are 60 girls and 90 boys born of the 150. Based on your study, do you believe that the percentage of girls born in China is 46.7?

A group of researchers research a common contagious disease. A newspaper found that 13 percent of Americans have been diagnosed with the disease in the last year. The researchers doubt that the percentage is really that high. It conducts its own survey. Out of 76 Americans surveyed, only two had been diagnosed with the disease. Would you agree with the newspaper's poll? In complete sentences, give three reasons why polls might give different results.

H₀: p ≥ 0.13
H_a: p < 0.13
Let P′ = the proportion of Americans who have the disease
normal for a single proportion
–2.688
p-value = 0.0036
Check student’s solution.
1. Alpha: 0.05
2. Decision: Reject the null hypothesis.
3. Reason for decision: The p-value is less than 0.05.
4. Conclusion: There is sufficient evidence to conclude that the percentage of Americans who have been diagnosed with the disease is less than 13 percent.
(0, 0.0623).
The plus-4s confidence interval is (0.0022, 0.0978)

The mean work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it’s shorter. She asks 10 engineering friends in start-ups for the lengths of their mean work weeks. Based on the results that follow, should she count on the mean work week to be shorter than 60 hours?

Data (length of mean work week): 70, 45, 55, 60, 65, 55, 55, 60, 50, 55.

Use the Lap time data for Lap 4 (see Appendix C: Data Sets) to test the claim that Terri finishes Lap 4, on average, in less than 129 seconds. Use all 20 races given.

H₀: μ ≥ 129
H_a: μ < 129
Let $\bar{X}$ = the average time in seconds that Terri finishes Lap 4.
Student's t-distribution
t = 1.209
0.8792
Check student’s solution.
1. Alpha: 0.05
2. Decision: Do not reject the null hypothesis.
3. Reason for decision: The p-value is greater than 0.05.
4. Conclusion: There is insufficient evidence to conclude that Terri’s mean lap time is less than 129 seconds.
(128.63, 130.37)

Use the Initial Public Offering data (see Appendix C: Data Sets) to test the claim that the mean offer price was $18 per share. Do not use all the data. Use your random number generator to randomly survey 15 prices.

"Asian Family Reunion," by Chau Nguyen

Every two years it comes around.

We all get together from different towns.

In my honest opinion,

It's not a typical family reunion.

Not forty, or fifty, or sixty,

But how about seventy companions!

The kids would play, scream, and shout

One minute they're happy, another they'll pout.

The teenagers would look, stare, and compare

From how they look to what they wear.

The men would chat about their business

That they make more, but never less.

Money is always their subject

And there's always talk of more new projects.

The women get tired from all of the chats

They head to the kitchen to set out the mats.

Some would sit and some would stand

Eating and talking with plates in their hands.

Then come the games and the songs

And suddenly, everyone gets along!

With all that laughter, it's sad to say

That it always ends in the same old way.

They hug and kiss and say "good-bye"

And then they all begin to cry!

I say that 60 percent shed their tears

But my mom counted 35 people this year.

She said that boys and men will always have their pride,

So we won't ever see them cry.

I myself don't think she's correct,

So could you please try this problem to see if you object?

H₀: p = 0.60
H_a: p < 0.60
Let P′ = the proportion of family members who shed tears at a reunion.
normal for a single proportion
–1.71
0.0438
Check student’s solution.
1. Alpha: 0.05
2. Decision: Reject the null hypothesis.
3. Reason for decision: p-value < alpha
4. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the proportion of family members who shed tears at a reunion is less than 0.60. However, the test is weak because the p-value and alpha are quite close, so other tests should be done.
We are 95 percent confident that between 38.29 percent and 61.71 percent of family members will shed tears at a family reunion. (0.3829, 0.6171). The plus-4s confidence interval (see chapter 8) is (0.3861, 0.6139)

Note that here the large-sample 1 – PropZTest provides the approximate p-value of 0.0438. Whenever a p-value based on a normal approximation is close to the level of significance, the exact p-value based on binomial probabilities should be calculated whenever possible. This is beyond the scope of this course.

"Blowing Bubbles," by Sondra Prull

Studying stats just made me tense,

I had to find some sane defense.

Some light and lifting simple play

To float my math anxiety away.

Blowing bubbles lifts me high

Takes my troubles to the sky.

POIK! They're gone, with all my stress

Bubble therapy is the best.

The label said each time I blew

The average number of bubbles would be at least 22.

I blew and blew and this I found

From 64 blows, they all are round!

But the number of bubbles in 64 blows

Varied widely, this I know.

20 per blow became the mean

They deviated by 6, and not 16.

From counting bubbles, I sure did relax

But now I give to you your task.

Was 22 a reasonable guess?

Find the answer and pass this test!

H₀: μ ≥ 22
H_a: μ < 22
Let $\bar{X}$ = the mean number of bubbles per blow.
Student's t-distribution
–2.667
p-value = 0.00486
Check student’s solution.
1. Alpha: 0.05
2. Decision: Reject the null hypothesis.
3. Reason for decision: The p-value is less than 0.05.
4. Conclusion: There is sufficient evidence to conclude that the mean number of bubbles per blow is less than 22.
(18.501, 21.499)

"Dalmatian Darnation," by Kathy Sparling

A greedy dog breeder named Spreckles

Bred puppies with numerous freckles

The Dalmatians he sought

Possessed spot upon spot

The more spots, he thought, the more shekels.

His competitors did not agree

That freckles would increase the fee.

They said, “Spots are quite nice

But they don't affect price;

One should breed for improved pedigree.”

The breeders decided to prove

This strategy was a wrong move.

Breeding only for spots

Would wreak havoc, they thought.

His theory they want to disprove.

They proposed a contest to Spreckles

Comparing dog prices to freckles.

In records they looked up

One hundred one pups:

Dalmatians that fetched the most shekels.

They asked Mr. Spreckles to name

An average spot count he'd claim

To bring in big bucks.

Said Spreckles, “Well, shucks,

It's for one hundred one that I aim.”

Said an amateur statistician

Who wanted to help with this mission.

“Twenty-one for the sample

Standard deviation's ample.”

They examined one hundred and one

Dalmatians that fetched a good sum.

They counted each spot,

Mark, freckle, and dot

And tallied up every one.

Instead of one hundred one spots

They averaged ninety-six dots

Can they muzzle Spreckles’

Obsession with freckles

Based on all the dog data they've got?

Macaroni and Cheese, please!! by Nedda Misherghi and Rachelle Hall

As a poor starving student I don't have much money to spend for even the bare necessities. So my favorite and main staple food is macaroni and cheese. It's high in taste and low in cost and nutritional value.

One day, as I sat down to determine the meaning of life, I got a serious craving for this, oh, so important, food of my life. So I went down the street to Greatway to get a box of macaroni and cheese, but it was SO expensive! $2.02 !!! Can you believe it? It made me stop and think. The world is changing fast. I had thought that the mean cost of a box (the normal size, not some super-gigantic-family-value-pack) was at most $1, but now I wasn't so sure. However, I was determined to find out. I went to 53 of the closest grocery stores and surveyed the prices of macaroni and cheese. Here are the data I wrote in my notebook:

Price per box of Mac and Cheese

5 stores @ $2.02
15 stores @ $0.25
3 stores @ $1.29
6 stores @ $0.35
4 stores @ $2.27
7 stores @ $1.50
5 stores @ $1.89
8 stores @ $0.75

I could see that the cost varied but I had to sit down to figure out whether or not I was right. If it does turn out that this mouth-watering dish is at most $1, then I'll throw a big cheesy party in our next statistics lab, with enough macaroni and cheese for just me. After all, as a poor starving student I can't be expected to feed our class of animals!

H₀: μ ≤ 1
H_a: μ > 1
Let $\bar{X}$ = the mean cost in dollars of macaroni and cheese in a certain town.
Student's t-distribution
t = 0.340
p-value = 0.36756
Check student’s solution.
1. Alpha: 0.05
2. Decision: Do not reject the null hypothesis.
3. Reason for decision: The p-value is greater than 0.05
4. Conclusion: The mean cost could be $1, or less. At the 5 percent significance level, there is insufficient evidence to conclude that the mean price of a box of macaroni and cheese is more than $1.
(0.8291, 1.241)

"William Shakespeare: The Tragedy of Hamlet, Prince of Denmark," by Jacqueline Ghodsi

THE CHARACTERS (in order of appearance):

HAMLET, Prince of Denmark and student of statistics
POLONIUS, Hamlet’s tutor
HORATIO, friend to Hamlet and fellow student

Scene: The great library of the castle, in which Hamlet does his lessons

Act I

The day is fair, but the face of Hamlet is clouded. He paces the large room. His tutor, Polonius, is reprimanding Hamlet regarding the latter’s recent experience. Horatio is seated at the large table at right stage.

POLONIUS: My Lord, how cans’t thou admit that thou hast seen a ghost! It is but a figment of your imagination!

HAMLET: I beg to differ; I know of a certainty that five-and-seventy in one hundred of us, condemned to the whips and scorns of time as we are, have gazed upon a spirit of health, or goblin damn’d, be their intents wicked or charitable.

POLONIUS: If thou dost insist upon thy wretched vision then let me invest your time; be true to thy work and speak to me through the reason of the null and alternate hypotheses. (He turns to Horatio.) Did not Hamlet himself say, “What a piece of work is man, how noble in reason, how infinite in faculties”? Then let not this foolishness persist. Go, Horatio, make a survey of three-and-sixty and discover what the true proportion be. For my part, I will never succumb to this fantasy, but deem man to be devoid of all reason should thy proposal of at least five-and-seventy in one hundred hold true.

HORATIO (to Hamlet): What should we do, my Lord?

HAMLET: Go to thy purpose, Horatio.

HORATIO: To what end, my Lord?

HAMLET: That you must teach me. But let me conjure you by the rights of our fellowship, by the consonance of our youth, but the obligation of our ever-preserved love, be even and direct with me, whether I am right or no.

Horatio exits, followed by Polonius, leaving Hamlet to ponder alone.

Act II

The next day, Hamlet awaits anxiously the presence of his friend, Horatio. Polonius enters and places some books upon the table just a moment before Horatio enters.

POLONIUS: So, Horatio, what is it thou didst reveal through thy deliberations?

HORATIO: In a random survey, for which purpose thou thyself sent me forth, I did discover that one-and-forty believe fervently that the spirits of the dead walk with us. Before my God, I might not this believe, without the sensible and true avouch of mine own eyes.

POLONIUS: Give thine own thoughts no tongue, Horatio. (Polonius turns to Hamlet.) But look to’t I charge you, my Lord. Come Horatio, let us go together, for this is not our test. (Horatio and Polonius leave together.)

HAMLET: To reject, or not reject, that is the question: whether ‘tis nobler in the mind to suffer the slings and arrows of outrageous statistics, or to take arms against a sea of data, and, by opposing, end them. (Hamlet resignedly attends to his task.)

(Curtain falls)

"Untitled," by Stephen Chen

I've often wondered how software is released and sold to the public. Ironically, I work for a company that sells products with known problems. Unfortunately, most of the problems are difficult to create, which makes them difficult to fix. I usually use the test program X, which tests the product, to try to create a specific problem. When the test program is run to make an error occur, the likelihood of generating an error is 1 percent.

So, armed with this knowledge, I wrote a new test program Y that will generate the same error that test program X creates, but more often. To find out if my test program is better than the original, so that I can convince the management that I'm right, I ran my test program to find out how often I can generate the same error. When I ran my test program 50 times, I generated the error twice. While this may not seem much better, I think that I can convince the management to use my test program instead of the original test program. Am I right?

H₀: p = 0.01
H_a: p > 0.01
Let P′ = the proportion of errors generated
Normal for a single proportion
2.13
0.0165
Check student’s solution.
1. Alpha: 0.05
2. Decision: Reject the null hypothesis.
3. Reason for decision: The p-value is less than 0.05.
4. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the proportion of errors generated is more than 0.01.
Confidence interval: (0, 0.094).
The plus-4s confidence interval is (0.004, 0.144).

"Japanese Girls’ Names"

by Kumi Furuichi

It used to be very typical for Japanese girls’ names to end with “ko.” The trend might have started around my grandmothers’ generation and its peak might have been around my mother’s generation. “Ko” means “child” in Chinese characters. Parents would name their daughters with “ko” attaching to other Chinese characters that have meanings that they want their daughters to become, such as Sachiko—happy child, Yoshiko—a good child, Yasuko—a healthy child, and so on.

However, I noticed recently that only two out of nine of my Japanese girlfriends at this school have names that end with “ko.” More and more, parents seem to have become creative, modernized, and, sometimes, westernized in naming their children.

I have a feeling that, while 70 percent or more of my mother’s generation would have names with “ko” at the end, the proportion has dropped among my peers. I wrote down all my Japanese friends’, ex-classmates’, coworkers’, and acquaintances’ names that I could remember. Following are the names. Some are repeats. Test to see if the proportion has dropped for this generation.

Ai, Akemi, Akiko, Ayumi, Chiaki, Chie, Eiko, Eri, Eriko, Fumiko, Harumi, Hitomi, Hiroko, Hiroko, Hidemi, Hisako, Hinako, Izumi, Izumi, Junko, Junko, Kana, Kanako, Kanayo, Kayo, Kayoko, Kazumi, Keiko, Keiko, Kei, Kumi, Kumiko, Kyoko, Kyoko, Madoka, Maho, Mai, Maiko, Maki, Miki, Miki, Mikiko, Mina, Minako, Miyako, Momoko, Nana, Naoko, Naoko, Naoko, Noriko, Rieko, Rika, Rika, Rumiko, Rei, Reiko, Reiko, Sachiko, Sachiko, Sachiyo, Saki, Sayaka, Sayoko, Sayuri, Seiko, Shiho, Shizuka, Sumiko, Takako, Takako, Tomoe, Tomoe, Tomoko, Touko, Yasuko, Yasuko, Yasuyo, Yoko, Yoko, Yoko, Yoshiko, Yoshiko, Yoshiko, Yuka, Yuki, Yuki, Yukiko, Yuko, Yuko.

"Phillip’s Wish," by Suzanne Osorio

My nephew likes to play

Chasing the girls makes his day.

He asked his mother

If it is okay

To get his ear pierced.

She said, “No way!”

To poke a hole through your ear,

Is not what I want for you, dear.

He argued his point quite well,

Says even my macho pal, Mel,

Has gotten this done.

It’s all just for fun.

C’mon please, mom, please, what the hell.

Again Phillip complained to his mother,

Saying half his friends (including their brothers)

Are piercing their ears

And they have no fears

He wants to be like the others.

She said, “I think it’s much less.

We must do a hypothesis test.

And if you are right,

I won’t put up a fight.

But, if not, then my case will rest.”

We proceeded to call fifty guys

To see whose prediction would fly.

Nineteen of the fifty

Said piercing was nifty

And earrings they’d occasionally buy.

Then there’s the other thirty-one,

Who said they’d never have this done.

So now this poem’s finished.

Will his hopes be diminished,

Or will my nephew have his fun?

H₀: p = 0.50
H_a: p < 0.50
Let P′ = the proportion of friends that has a pierced ear.
normal for a single proportion
–1.70
p-value = 0.0448
Check student’s solution.
1. Alpha: 0.05
2. Decision: Reject the null hypothesis.
3. Reason for decision: The p-value is less than 0.05. (However, they are very close.)
4. Conclusion: There is sufficient evidence to support the claim that less than 50 percent of his friends have pierced ears.
Confidence interval: (0.245, 0.515): The plus-4s confidence interval is (0.259, 0.519).

"The Craven," by Mark Salangsang

Once upon a morning dreary

In stats class I was weak and weary.

Pondering over last night’s homework

Whose answers were now on the board

This I did and nothing more.

While I nodded nearly napping

Suddenly, there came a tapping.

As someone gently rapping,

Rapping my head as I snore.

Quoth the teacher, “Sleep no more.”

“In every class you fall asleep,”

The teacher said, his voice was deep.

“So a tally I’ve begun to keep

Of every class you nap and snore.

The percentage being forty-four.”

“My dear teacher I must confess,

While sleeping is what I do best.

The percentage, I think, must be less,

A percentage less than forty-four.”

This I said and nothing more.

“We’ll see,” he said and walked away,

And fifty classes from that day

He counted till the month of May

The classes in which I napped and snored.

The number he found was twenty-four.

At a significance level of 0.05,

Please tell me am I still alive?

Or did my grade just take a dive

Plunging down beneath the floor?

Upon thee I hereby implore.

Toastmasters International cites a report by Gallup Poll that 40 percent of Americans fear public speaking. A student believes that less than 40 percent of students at her school fear public speaking. She randomly surveys 361 schoolmates and finds that 135 report they fear public speaking. Conduct a hypothesis test to determine if the percentage at her school is less than 40.

H₀: p = 0.40
H_a: p < 0.40
Let P′ = the proportion of schoolmates who fear public speaking.
normal for a single proportion
–1.01
p-value = 0.1563
Check student’s solution.
1. Alpha: 0.05
2. Decision: Do not reject the null hypothesis.
3. Reason for decision: The p-value is greater than 0.05.
4. Conclusion: There is insufficient evidence to support the claim that less than 40 percent of students at the school fear public speaking.
Confidence interval: (0.3241, 0.4240): The plus-4s confidence interval is (0.3257, 0.4250).

Sixty-eight percent of online courses taught at community colleges nationwide were taught by full-time faculty. To test if 68 percent also represents California’s percent for full-time faculty teaching the online classes, Long Beach City College (LBCC) in California was randomly selected for comparison. In the same year, 34 of the 44 online courses LBCC offered were taught by full-time faculty. Conduct a hypothesis test to determine if 68 percent represents California. Note: For more accurate results, use more California community colleges and this past year's data.

According to an article in a local poll, a city found that 14 percent of its residents walk for exercise. Suppose that a survey is conducted to determine this years rate. Nine out of 70 randomly chosen city residents replied that they walk for exercise. Conduct a hypothesis test to determine if the rate is still 14 percent or if it has decreased.

H₀: p = 0.14
H_a: p < 0.14
Let P′ = the proportion of nursing home residents that have the disease.
normal for a single proportion
–0.2756
p-value = 0.3914
Check student’s solution.
1. Alpha: 0.05
2. Decision: Do not reject the null hypothesis.
3. Reason for decision: The p-value is greater than 0.05.
4. At the 5 percent significance level, there is insufficient evidence to conclude that the proportion of nursing home residents that have the disease is less than 0.14.
Confidence interval: (0.0502, 0.2070): The plus-4s confidence interval (see chapter 8) is (0.0676, 0.2297).

The mean age of De Anza College students in a previous term was 26.6 years old. An instructor thinks the mean age for online students is older than 26.6. She randomly surveys 56 online students and finds that the sample mean is 29.4 with a standard deviation of 2.1. Conduct a hypothesis test.

Registered nurses earned an average annual salary of $69,110. For that same year, a survey was conducted of 41 California registered nurses to determine if the annual salary is higher than $69,110 for California nurses. The sample average was $71,121 with a sample standard deviation of $7,489. Conduct a hypothesis test.

H₀: μ = 69,110
H_a: μ > 69,110
Let $\bar{X}$ = the mean salary in dollars for California registered nurses.
Student's t-distribution
t = 1.719
p-value: 0.0466
Check student’s solution.
1. Alpha: 0.05
2. Decision: Reject the null hypothesis.
3. Reason for decision: The p-value is less than 0.05.
4. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the mean salary of California registered nurses exceeds $69,110.
($68,757, $73,485)

La Leche League International reports that the mean age of weaning a child from breastfeeding is age four to five worldwide. In America, most nursing mothers wean their children much earlier. Suppose a random survey is conducted of 21 U.S. mothers who recently weaned their children. The mean weaning age was nine months (3/4 year) with a standard deviation of 4 months. Conduct a hypothesis test to determine if the mean weaning age in the United States is less than four years old.

Harley Davidson motorcycles are the largest selling motorcycle in the United States, with 14 percent of all motorcycles sold in 2012. Interestingly, a random sample of 1,945 stolen motorcycles was selected, and it was found that just 8 percent of them were Harleys. Is there good evidence that the proportion of Harleys among stolen motorcycles is significantly less than their share of all motorcycles? After conducting the test, what decision and conclusion would you make?

Reject H₀: There is sufficient evidence to conclude that the proportion of Harleys stolen is significantly less than their share of all motorcycles
Do not reject H₀: There is not sufficient evidence to conclude that the proportion of Harleys stolen is significantly less than their share of all motorcycles
Do not reject H₀: There is sufficient evidence to conclude that the proportion of Harleys stolen is significantly more than their share of all motorcycles
Reject H₀: There is sufficient evidence to conclude that the proportion of Harleys stolen is significantly more than their share of all motorcycles

H₀: p ≥ 0.14, H_a: p < 0.14
p-value < 0.0002
Alpha: 0.05
Reject the null hypothesis.
At the 5 percent significance level, there is sufficient evidence to conclude that the proportion of Harleys stolen is significantly less than their share of all motorcycles. (conclusion a)

A statistics instructor believes that fewer than 20 percent of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing.
At a 1 percent level of significance, what is an appropriate conclusion?

There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20 percent.
There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is more than 20 percent.
There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20 percent.
There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is at least 20 percent.

Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test.
At a significance level of a = 0.05, what is the correct conclusion?

There is enough evidence to conclude that the mean number of hours is more than 4.75.
There is enough evidence to conclude that the mean number of hours is more than 4.5.
There is not enough evidence to conclude that the mean number of hours is more than 4.5.
There is not enough evidence to conclude that the mean number of hours is more than 4.75.

Hypothesis testing: For the following 10 exercises, answer each question.

State the null and alternate hypotheses.
State the p-value.
State alpha.
What is your decision?
Write a conclusion.
Answer any other questions asked in the problem.

A research group is studying a particular infectious disease. In 2011 at least 18 percent of nursing home residents had the disease. An Introduction to Statistics class in Daviess County, KY, conducted a hypothesis test at the nursing home (approximately 1,200 residents) to determine if the local nursing home's incidence was lower. One hundred fifty residents were chosen at random and surveyed. Of the 150 residents surveyed, 82 have the disease. Use a significance level of 0.05 and, using appropriate statistical evidence, conduct a hypothesis test and state the conclusions.

A recent survey in the New York Times Almanac indicated that 48.8 percent of families own stock. A broker wanted to determine if this survey could be valid. He surveyed a random sample of 250 families and found that 142 owned some type of stock. At the 0.05 significance level, can the survey be considered to be accurate?

H₀: p = 0.488 H_a: p ≠ 0.488
p-value = 0.0114
alpha = 0.05
Reject the null hypothesis.
At the 5 percent level of significance, there is enough evidence to conclude that 48.8 percent of families own stocks.
The survey does not appear to be accurate.

Driver error can be listed as the cause of approximately 54 percent of all fatal auto accidents, according to the American Automobile Association. Thirty randomly selected fatal accidents are examined, and it is determined that 14 were caused by driver error. Using α = 0.05, is the AAA proportion accurate?

The U.S. Department of Energy reported that 51.7 percent of homes were heated by natural gas. A random sample of 221 homes in Kentucky found that 115 were heated by natural gas. Does the evidence support the claim for Kentucky at the α = 0.05 level? Are the results applicable across the country? Why?

H₀: p = 0.517 H_a: p ≠ 0.517
p-value = 0.9203.
alpha = 0.05.
Do not reject the null hypothesis.
At the 5 percent significance level, there is not enough evidence to conclude that the proportion of homes in Kentucky that are heated by natural gas is 0.517.
However, we cannot generalize this result to the entire nation. First, the sample’s population is only the state of Kentucky. Second, it is reasonable to assume that homes in the extreme north and south will have extreme high usage and low usage, respectively. We would need to expand our sample base to include these possibilities if we wanted to generalize this claim to the entire nation.

For Americans using library services, the American Library Association claims that at most 67 percent of patrons borrow books. The library director in Owensboro, KY, feels this is not true, so she asked a local college statistic class to conduct a survey. The class randomly selected 100 patrons and found that 82 borrowed books. Did the class demonstrate that the percentage was higher in Owensboro, KY? Use α = 0.01 level of significance. What is the possible proportion of patrons who do borrow books from the Owensboro Library?

The Weather Underground reported that the mean amount of summer rainfall for the northeastern United States is at least 11.52 inches. Ten cities in the northeast are randomly selected and the mean rainfall amount is calculated to be 7.42 inches with a standard deviation of 1.3 inches. At the α = 0.05 level, can it be concluded that the mean rainfall was below the reported average? What if α = 0.01? Assume the amount of summer rainfall follows a normal distribution.

H₀: µ ≥ 11.52 H_a: µ < 11.52
p-value = 0.000002 which is almost 0.
alpha = 0.05.
Reject the null hypothesis.
At the 5 percent significance level, there is enough evidence to conclude that the mean amount of summer rain in the northeaster US is less than 11.52 inches, on average.
We would make the same conclusion if alpha was 1 percent because the p-value is almost 0.

A survey in the New York Times Almanac finds the mean commute time (one way) is 25.4 minutes for the 15 largest US cities. The Austin, TX, chamber of commerce feels that Austin’s commute time is less and wants to publicize this fact. The mean for 25 randomly selected commuters is 22.1 minutes with a standard deviation of 5.3 minutes. At the α = 0.10 level, is the Austin, TX, commute significantly less than the mean commute time for the 15 largest U.S. cities?

A report by the Gallup Poll found that a woman visits her doctor, on average, at most 5.8 times each year. A random sample of 20 women results in these yearly visit totals:

At the α = 0.05 level, can it be concluded that the sample mean is higher than 5.8 visits per year?

H₀: µ ≤ 5.8 H_a: µ > 5.8
p-value = 0.9987
alpha = 0.05
Do not reject the null hypothesis.
At the 5 percent level of significance, there is not enough evidence to conclude that a woman visits her doctor, on average, more than 5.8 times a year.

According to the New York Times Almanac the mean family size in the United States is 3.18. A sample of a college math class resulted in the following family sizes:

At α = 0.05, is the class’s mean family size greater than the national average? Does the Almanac result remain valid? Why?

The student academic group on a college campus claims that freshman students study at least 2.5 hours per day, on average. One Introduction to Statistics class was skeptical. The class took a random sample of 30 freshman students and found a mean study time of 137 minutes with a standard deviation of 45 minutes. At α = 0.01 level, is the student academic group’s claim correct?

H₀: µ ≥ 150 H_a: µ < 150
p-value = 0.0622
alpha = 0.01
Do not reject the null hypothesis.
At the 1 percent significance level, there is not enough evidence to conclude that freshmen students study less than 2.5 hours per day, on average.
The student academic group’s claim appears to be correct.

References

American Automobile Association. (n.d.). Retrieved from www.aaa.com

American Library Association. (n.d.). Retrieved from www.ala.org

Amit Schitai. (n.d.). Data.

Bureau of Labor Statistics. (n.d.). Occupational employment statistics. Retrieved from http://www.bls.gov/oes/current/oes291111.htm

Centers for Disease Control and Prevention. (n.d.). Retrieved from www.cdc.gov

De Anza College. (2006). Foothill-De Anza Community College District. Retrieved from http://research.fhda.edu/factbook/DAdemofs/Fact_sheet_da_2006w.pdf

Federal Bureau of Investigation. (n.d.). Uniform crime reports and index of crime in Daviess in the state of Kentucky enforced by Daviess County from 1985 to 2005. Retrieved from http://www.disastercenter.com/kentucky/crime/3868.htm

Gallup. (n.d.). Retrieved from www.gallup.com

Johansen, C., Boice, Jr. J., McLaughlin, J., & Olsen, J. (2001, Feb. 7). Cellular telephones and cancera nationwide cohort study in Denmark. Journal of National Cancer Institute, 93(3), 2037. Retrieved from http://www.ncbi.nlm.nih.gov/pubmed/11158188

La Leche League International. (n.d.). Retrieved from http://www.lalecheleague.org/Law/BAFeb01.html

Online Learning Consortium. (2005 Nov.). Growing by degrees: Online education in the United States, 2005. Newburyport, MA: Allen, I. E., & Seaman, J. Available at http://files.eric.ed.gov/fulltext/ED530062.pdf

Toastmasters International. (n.d.). Retrieved from http://toastmasters.org/artisan/detail.asp?CategoryID=1&SubCategoryID=10&ArticleID=429&Page=1

U.S. Census Bureau. (n.d.). Language use. Retrieved from https://www.census.gov/topics/population/language-use.html

U.S. Census Bureau. (n.d.). QuickFacts. Retrieved from https://www.census.gov/quickfacts/table/PST045216/00

U.S. Department of Energy. (n.d.). Retrieved from http://energy.gov

Weather Underground. (n.d.). Retrieved from www.wunderground.com

Attribution. Statistics, OpenStax, Rice University, and the contributing authors credited in the book. Licensed under Creative Commons Attribution 4.0 International. Source revision 7dea80ae28 of openstax/osbooks-statistics (2026-04-23). CC BY 4.0. Original source ↗. Text edition of a pinned CC BY 4.0 source revision of the OpenStax book, converted from CNXML by the EvidenceHelix importer. Original prose, learning objectives, tables, exercises, glossary and captions are preserved; images are not copied (figures link to the original source and keep their credits). No new content was added. Later revisions of this book can carry different licence terms and are not used.