Comparing Two Independent Population Proportions

Two types of medication for hives are being tested to determine if there is a difference in the proportions of adult patient reactions. Twenty out of a random sample of 200 adults given Medication A still had hives 30 minutes after taking the medication. Twelve out of another random sample of 200 adults given Medication B still had hives 30 minutes after taking the medication. Test at a 1 percent level of significance.

The problem asks for a difference in proportions, making it a test of two proportions.

Let A and B be the subscripts for Medication A and Medication B, respectively. Then, p_A and p_B are the desired population proportions.

Random Variable:

H₀: p_A = p_B

p_A – p_B = 0

H_a: p_A ≠ p_B

p_A – p_B ≠ 0

The words is a difference tell you the test is two-tailed.

Distribution for the test: Since this is a test of two binomial population proportions, the distribution is normal:

$p_c=\frac{x_A+x_B}{n_A+n_B}=\frac{20+12}{200+200}=0.081–p_c=0.92$

${P^{\prime }}_A–{P^{\prime }}_B~N\left[0,\sqrt{\left(0.08\right)\left(0.92\right)(\frac{1}{200}+\frac{1}{200})}\right]$

P′_A – P′_B follows an approximate normal distribution.

Calculate the p-value using the normal distribution: p-value = 0.1404.

Estimated proportion for group A: ${p^{\prime }}_A=\frac{x_A}{n_A}=\frac{20}{200}=0.1$

Estimated proportion for group B: ${p^{\prime }}_B=\frac{x_B}{n_B}=\frac{12}{200}=0.06$

Graph:

P′_A – P′_B = 0.1 – 0.06 = 0.04.

Half the p-value is below –0.04, and half is above 0.04.

Compare α and the p-value: α = 0.01 and the p-value = 0.1404. α < p-value.

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

Conclusion: At a 1 percent level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the proportions of adult patients who did not react after 30 minutes to Medication A and Medication B.

A research study was conducted about gender differences in texting. The researcher believed that the proportion of girls involved in texting is less than the proportion of boys involved. The data collected in spring 2010 among a random sample of middle and high school students in a large school district in the southern United States is summarized in fs-idm29483888. Is the proportion of girls sending texts less than the proportion of boys texting? Test at a 1 percent level of significance.

This is a test of two population proportions. Let M and F be the subscripts for males and females. Then, p_M and p_F are the desired population proportions.

Random variable:

H₀: p_F = p_M H₀: p_F – p_M = 0

H_a: p_F < p_M H_a: p_F – p_M < 0

The words less than tell you the test is left-tailed.

Distribution for the test: Since this is a test of two population proportions, the distribution is normal:

$p_c=\frac{x_F+x_M}{n_F+n_M}=\frac{156+183}{2169+2231}=\text{0}\text{.077}$
$1-p_c=0.923$
Therefore,
${p^{\prime }}_F–{p^{\prime }}_M\sim N\left(0,\sqrt{\left(0.077\right)\left(0.923\right)\left(\frac{1}{2169}+\frac{1}{2231}\right)}\right)$
p′_F – p′_M follows an approximate normal distribution.

Calculate the p-value using the normal distribution:
p-value = 0.1045
Estimated proportion for females: 0.0719
Estimated proportion for males: 0.082

Graph:

Decision: Since α < p-value, do not reject H₀.

Conclusion: At the 1 percent level of significance, from the sample data, there is not sufficient evidence to conclude that the proportion of girls sending texts is less than the proportion of boys sending texts.

Researchers conducted a study of smartphone use (Phone A versus Phone B) among adults. A cell phone company claimed that Phone B smartphones are more popular with whites (non-Hispanic) than with African Americans. The results of the survey indicate that of the 232 African American cell phone owners randomly sampled, 5 percent own Phone B. Of the 1,343 white cell phone owners randomly sampled, 10 percent own Phone B. Test at the 5 percent level of significance. Is the proportion of white Phone B owners greater than the proportion of African American Phone B owners?

This is a test of two population proportions. Let W and A be the subscripts for the whites and African Americans. Then, p_W and p_A are the desired population proportions.

Random variable:

H₀: p_W = p_A H₀: p_W – p_A = 0

H_a: p_W > p_A H_a: p_W – p_A > 0

The words more popular indicate that the test is right-tailed.

Distribution for the test: The distribution is approximately normal.

$p_c=\frac{x_W+x_A}{n_W+n_A}=\frac{134+12}{1343+232}=0.0927$

$1-p_c=0.9073$

Therefore,

${p^{\prime }}_W–{p^{\prime }}_A\backsim N\left(0,\sqrt{\left(0.0927\right)\left(0.9073\right)\left(\frac{1}{1343}+\frac{1}{232}\right)}\right)$

${p^{\prime }}_W–{p^{\prime }}_A$ follows an approximate normal distribution.

Calculate the p-value using the normal distribution:

Graph:

Decision: Since α > p-value, reject the H₀.

Conclusion: At the 5 percent level of significance, from the sample data, there is sufficient evidence to conclude that a larger proportion of white cell phone owners use Phone B than African Americans.

Is this a test of means or proportions?

What is the random variable?

P′_OS1 – P′_OS2 = difference in the proportions of phones that had system failures within the first eight hours of operation with OS₁ and OS₂.

State the null and alternative hypotheses.

What is the p-value?

0.1018

What can you conclude about the two operating systems?

Is this a test of means or proportions?

proportions

State the null and alternative hypotheses.

1. H₀: _________
2. H_a: _________

Is this a right-tailed, left-tailed, or two-tailed test? How do you know?

right-tailed

What is the random variable of interest for this test?

In words, define the random variable for this test.

The random variable is the difference in proportions (percents) of the populations that are of two or more races in Nevada and North Dakota.

Which distribution (normal or Student’s t) would you use for this hypothesis test?

Explain why you chose the distribution you did for the Exercise 10.56.

Our sample sizes are much greater than five each, so we use the normal for two proportions distribution for this hypothesis test.

Calculate the test statistic.

Sketch a graph of the situation. Mark the hypothesized difference and the sample difference. Shade the area corresponding to the p-value.

Check student’s solution.

Find the p-value.

At a preconceived α = 0.05, write the following:

1. Your decision:
2. The reason for your decision:
3. Your conclusion (write out in a complete sentence):

1. Reject the null hypothesis.
2. p-value < alpha
3. At the 5 percent significance level, there is sufficient evidence to conclude that the proportion (percent) of the population that is of two or more races in Nevada is statistically higher than that in North Dakota.

Does it appear that the proportion of Nevadans who are two or more races is higher than the proportion of North Dakotans? Why or why not?

A recent drug survey showed an increase in the use of prescription medication among local senior citizens as compared to the national percent. Suppose that a survey of 100 local seniors and 100 national seniors is conducted to see if the proportion of prescription medication use is higher locally or nationally. Locally, 65 senior citizens reported taking prescription medication within the past month, while 60 national seniors reported using them.

Elizabeth Mjelde, an art history professor, was interested in whether the value from the Golden Ratio formula, $\left(\frac{\text{larger + smaller dimension}}{\text{larger dimension}}\right)$, was the same in the Whitney Exhibit for works from 1900 to 1919 as for works from 1920 to 1942. Thirty-seven early works were sampled, averaging 1.74 with a standard deviation of 0.11. Sixty-five of the later works were sampled, averaging 1.746 with a standard deviation of 0.1064. Do you think that there is a significant difference in the Golden Ratio calculation?

A year was randomly picked from 1985 to the present. In that year, there were 2,051 Hispanic students at Cabrillo College out of a total of 12,328 students. At Lake Tahoe College, there were 321 Hispanic students out of a total of 2,441 students. In general, do you think that the percent of Hispanic students at the two colleges is basically the same or different?

Subscripts: 1 = Cabrillo College, 2 = Lake Tahoe College

1. H₀: p₁ = p₂
2. H_a: p₁ ≠ p₂
3. The random variable is the difference between the proportions of Hispanic students at Cabrillo College and Lake Tahoe College.
4. normal for two proportions
5. test statistic: 4.29
6. p-value: 0.00002
7. Check student’s solution.
8. 1. Alpha: 0.05
   2. Decision: Reject the null hypothesis.
   3. Reason for decision: p-value < alpha
   4. Conclusion: There is sufficient evidence to conclude that the proportions of Hispanic students at Cabrillo College and Lake Tahoe College are different.

This is

1. a test of two proportions
2. a test of two independent means
3. a test of a single mean
4. a test of matched pairs.

An appropriate null hypothesis is

1. p₂₀₁₁ ≤ p₂₀₁₀
2. p₂₀₁₁ ≥ p₂₀₁₀
3. μ₂₀₁₁ ≤ μ₂₀₁₀
4. p₂₀₁₁ > p₂₀₁₀

a

The p-value is 0.0022. At a 1 percent level of significance, what is the appropriate conclusion?

1. There is sufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile virus is less than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile virus.
2. There is insufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile virus is more than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile virus.
3. There is insufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile virus is less than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile virus.
4. There is sufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile virus is more than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile virus.

Researchers conducted a study to find out if there is a difference in the use of e-readers by different age groups. Randomly selected participants were divided into two age groups. In the 16- to 29-year-old group, 7 percent of the 628 surveyed use e-readers, while 11 percent of the 2,309 participants 30 years old and older use e-readers.

Test: two independent sample proportions.

Random variable: p′₁ - p′₂

Distribution:
H₀: p₁ = p₂
H_a: p₁ ≠ p₂

The proportion of e-reader users is different for the 16- to 29-year-old users from that of the 30 and older users.

Graph: two-tailed

p-value : 0.0033

Decision: Reject the null hypothesis.

Conclusion: At the 5 percent level of significance, from the sample data, there is sufficient evidence to conclude that the proportion of e-reader users 16 to 29 years old is different from the proportion of e-reader users 30 and older.

Adults aged 18 years and older were randomly selected for a survey about a specific disease. The researchers wanted to determine if the proportion of women who have the disease is less than the proportion of southern men who do. The results are shown in fs-idp62517488. Test at the 1 percent level of significance.

Two computer users were discussing tablet computers. A higher proportion of people ages 16 to 29 use tablets than of people age 30 and older. fs-idp42650880 details the number of tablet owners for each age group. Test at the 1 percent level of significance.

Test: two independent sample means, population standard deviations unknown.

${\mu }_1$ = the the mean price of a sociology text on the selected site.

${\mu }_2$ = the mean price of a math/science text on the selected site.

Random variable: $\overline{X_1}-\overline{X_1}$ = the difference in the sample mean textbook price between sociology texts and math/science texts.

Hypotheses: $H_0 : {\mu }_1-{\mu }_2 = 0, H_a : {\mu }_1 - {\mu }_2 < {\mu }_2$ which can be expressed as $H0s: \mu 1-\mu 2, \mathrm{Ha} \mu 1 < \mu 2$.

Distribution for the test: Use $t_{df}$; because each sample has more than 30 observations, $df=n_1+n_2-2=33+33-2=64$.

Estimate the critical value on the $t$-table using the nearest available degrees of freedom, 60. The critical value, 2.660, is found in the .0005 column.

Calculate the test statistic: $t_c=\frac{({\overline{X}}_1-{\overline{X}}_2)-0}{\sqrt{\frac{{s_1}^2}{n_2}+\frac{{s_2}^2}{n_2}}}=\frac{(74.64-111.56)-0}{\sqrt{\frac{49.{36}^2}{33}+\frac{66.{90}^2}{33}}}=-2.55$.

Using a calculator with $t_c=-2.55$ and $df=64$, the left-tailed $p$-value: Decision: Reject $H_0$. Conclusion: At the 1% level of significance, from the sample data, there is sufficient evidence to conclude that the mean price of sociology textbooks is less than the mean price of textbooks for math/science.

A group of friends debated whether more men use smartphones than women. They consulted a research study of smartphone use among adults. The results of the survey indicate that of the 973 men randomly sampled, 379 use smartphones. For women, 404 of the 1,304 who were randomly sampled use smartphones. Test at the 5 percent level of significance.

While her husband spent 2.5 hours picking out new speakers, a statistician decided to determine whether the percent of men who enjoy shopping for electronic equipment is higher than the percent of women who do. The population was Saturday afternoon shoppers. Out of 67 men, 24 said they enjoyed the activity. Eight of the 24 women surveyed claimed to enjoy the activity. Interpret the results of the survey.

Subscripts: 1: men; 2: women

1. H₀: p₁ ≤ p₂
2. H_a: p₁ > p₂
3. ${P^{\prime }}_1-{P^{\prime }}_2$ is the difference between the proportions of men and women who enjoy shopping for electronic equipment.
4. normal for two proportions
5. test statistic: 0.22
6. p-value: 0.4133
7. Check student’s solution.
8. 1. Alpha: 0.05
   2. Decision: Do not reject the null hypothesis.
   3. Reason for Decision: p-value > alpha
   4. Conclusion: At the 5 percent significance level, there is insufficient evidence to conclude that the proportion of men who enjoy shopping for electronic equipment is more than the proportion of women.

We are interested in whether children’s educational computer software costs less, on average, than children’s entertainment software. Thirty-six educational software titles were randomly picked from a catalog. The mean cost was $31.14 with a standard deviation of $4.69. Thirty-five entertainment software titles were randomly picked from the same catalog. The mean cost was $33.86 with a standard deviation of $10.87. Decide whether children’s educational software costs less, on average, than children’s entertainment software.

A researcher recently claimed that the proportion of college-age males who wear at least one piece of jewelery is as high as the proportion of college-age females. She conducted a survey in her classes. Out of 107 males, 20 wear at least one piece of jewelery. Out of 92 females, 47 wear at least one piece of jewelery. Do you believe that the proportion of males has reached the proportion of females?

1. H₀: p₁ = p₂
2. H_a: p₁ ≠ p₂
3. ${P^{\prime }}_1-{P^{\prime }}_2$ is the difference between the proportions of men and women that have at least one pierced ear.
4. normal for two proportions
5. test statistic: –4.82
6. p-value: zero
7. Check student’s solution.
8. 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 proportions of males and females with at least one pierced ear is different.

Use the data sets found in Appendix C to answer this exercise. Is the proportion of race laps Terri completes slower than 130 seconds less than the proportion of practice laps she completes slower than 135 seconds?

To Breakfast or Not to Breakfast? by Richard Ayore

In the American society, birthdays are one of those days that everyone looks forward to. People of different ages and peer groups gather to mark the 18th, 20th, …, birthdays. During this time, one looks back to see what he or she has achieved for the past year and also focuses ahead for more to come.

If, by any chance, I am invited to one of these parties, my experience is always different. Instead of dancing around with my friends while the music is booming, I get carried away by memories of my family back home in Kenya. I remember the good times I had with my brothers and sister while we did our daily routine.

Every morning, I remember we went to the shamba (garden) to weed our crops. I remember one day arguing with my brother as to why he always remained behind just to join us an hour later. In his defense, he said that he preferred waiting for breakfast before he came to weed. He said, “This is why I always work more hours than you guys!”

And so, to prove him wrong or right, we decided to give it a try. One day we went to work as usual without breakfast, and recorded the time we could work before getting tired and stopping. On the next day, we all ate breakfast before going to work. We recorded how long we worked again before getting tired and stopping. Of interest was our mean increase in work time. Though not sure, my brother insisted that it was more than two hours. Using the data in fs-idm167762016, solve our problem.

1. H₀: µ_d = 0
2. H_a: µ_d > 0
3. The random variable X_d is the mean difference in work times on days when eating breakfast and on days when not eating breakfast.
4. t₉
5. test statistic: 4.8963
6. p-value: 0.0004
7. Check student’s solution.
8. 1. Alpha: 0.05
   2. Decision: Reject the null hypothesis.
   3. Reason for Decision: p-value < alpha
   4. Conclusion: At the 5 percent level of significance, there is sufficient evidence to conclude that the mean difference in work times on days when eating breakfast and on days when not eating breakfast has increased.