Facts About the F Distribution

Let’s return to the slicing tomato exercise in fs-idp19091024. The means of the tomato yields under the five mulching conditions are represented by μ₁, μ₂, μ₃, μ₄, μ₅. We will conduct a hypothesis test to determine if all means are the same or at least one is different. Using a significance level of 5 percent, test the null hypothesis that there is no difference in mean yields among the five groups against the alternative hypothesis that at least one mean is different from the rest.

The null and alternative hypotheses are as follows:

H₀: μ₁ = μ₂ = μ₃ = μ₄ = μ₅

H_a: μ_i ≠ μ_j for some i ≠ j

The one-way ANOVA results are shown in eip-632

Distribution for the test: F_4,10

df(num) = 5 – 1 = 4

df(denom) = 15 – 5 = 10

Test statistic: F = 4.4810

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

Original figure description: This graph shows a nonsymmetrical F distribution curve. The horizontal axis extends from 0 - 5, and the vertical axis ranges from 0 - 0.7. The curve is strongly skewed to the right.

Probability statement: p-value = P(F > 4.481) = 0.0248

Compare α and the p-value: α = 0.05, p-value = 0.0248

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

Conclusion: At the 5 percent significance level, we have reasonably strong evidence that differences in mean yields for slicing tomato plants grown under different mulching conditions are unlikely to be due to chance alone. We may conclude that at least some of the mulches led to different mean yields.

Using a significance level of 1 percent, is there a difference in mean grades among the sororities?

Let μ₁, μ₂, μ₃, μ₄ be the population means of the sororities. Remember that the null hypothesis claims that the sorority groups are from the same normal distribution. The alternate hypothesis says that at least two of the sorority groups come from populations with different normal distributions. Notice that the four sample sizes are each five.

H₀: μ₁ = μ₂ = μ₃ = μ₄

H_a: Not all of the means μ₁, μ₂, μ₃, μ₄ are equal.

Distribution for the test: F_3,16

where k = 4 groups and n = 20 samples in total.

df(num)= k – 1 = 4 – 1 = 3

df(denom) = n – k = 20 – 4 = 16

Calculate the test statistic: F = 2.23

Graph

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

Original figure description: This graph shows a nonsymmetrical F distribution curve with values of 0 and 2.23 on the x-axis representing the test statistic of sorority grade averages. The curve is slightly skewed to the right, but is approximately normal. A vertical upward line extends from 2.23 to the curve and the area to the right of this is shaded to represent the p-value.

Probability statement: p-value = P(F > 2.23) = 0.1241

Compare α and the p-value: α = 0.01
p-value = 0.1241
α < p-value

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

Conclusion: There is not sufficient evidence to conclude that there is a difference among the mean grades for the sororities.

Does it appear that the three soils in which the bean plants were grown produce the same mean height? Test at a 3 percent level of significance.

This time, we will perform the calculations that lead to the F' statistic. Notice that each group has the same number of plants, so we will use the formula F' = $\frac{n\cdot s_{\overline{x}}{}^2}{s^2{}_{\text{pooled}}}$ .

First, calculate the sample mean and sample variance of each group.

Next, calculate the variance of the three group means by calculating the variance of 24.2, 25.4, and 24.4. Variance of the group means = 0.413 = $s_{\overline{x}}{}^2$,

then MS_between = $n{s}_{\overline{x}}{}^2$ = (5)(0.413) where n = 5 is the sample size (number of plants each child grew).

Calculate the mean of the three sample variances (11.7, 18.3, and 16.3). Mean of the sample variances = 15.433 = s²_pooled,

then MS_within = s²_pooled = 15.433.

The F statistic (or F ratio) is $F=\frac{M{S}_{\text{between}}}{M{S}_{\text{within}}}=\frac{n{s}_{\overline{x}}{}^2}{s^2{}_{pooled}}=\frac{\left(5\right)\left(0.413\right)}{15.433}=0.134\text{.}$

The dfs for the numerator = the number of groups – 1 = 3 – 1 = 2.

The dfs for the denominator = the total number of samples – the number of groups = 15 – 3 = 12.

The distribution for the test is F_2,12 and the F statistic is F = 0.134.

The p-value is P(F > 0.134) = 0.8759.

Decision: Since α = 0.03 and the p-value = 0.8759, we do not reject H₀. Why?

Conclusion: With a 3 percent level of significance from the sample data, the evidence is not sufficient to conclude that the mean heights of the bean plants are different.

An F statistic can have what values?

What happens to the curves as the degrees of freedom for the numerator and the denominator get larger?

The curves approximate the normal distribution.

What is the df(num)?

What is the df(denom)?

10

What are the sum of squares and mean squares factors?

What are the sum of squares and mean squares errors?

SS = 237.33; MS = 23.73

What is the F statistic?

What is the p-value?

0.1614

At the 5 percent significance level, is there a difference in the mean jump heights among the teams?

What is the df(num)?

two

What is the df(denom)?

What are the SS_between and MS_between?

SS = 5,700.4;

MS = 2,850.2

What are the SS_within and MS_within?

What is the F Statistic?

3.6101

What is the p-value?

At the 10 percent significance level, are the scores among the different groups different?

Yes, there is enough evidence to show that the scores among the groups are statistically significant at the 10 percent level.

p-value = ______

α = 0.05

a. Decision: ____________________________

b. Conclusion: ____________________________

α = 0.01

a. Decision: ____________________________

b. Conclusion: ____________________________

Three students, Linda, Tuan, and Javier, are given five laboratory rats each for a nutritional experiment. Each rat’s weight is recorded in grams. Linda feeds her rats Formula A, Tuan feeds his rats Formula B, and Javier feeds his rats Formula C. At the end of a specified time period, each rat is weighed again, and the net gain in grams is recorded. Using a significance level of 10 percent, test the hypothesis that the three formulas produce the same mean weight gain.

1. H₀: µ_L = µ_T = µ_J
2. H_a: at least any two of the means are different
3. df(num) = 2; df(denom) = 12
4. F distribution
5. 0.67
6. 0.5305
7. Check student’s solution.
8. Decision: Do not reject null hypothesis.
9. Conclusion: There is insufficient evidence to conclude that the means are different.

A grassroots group opposed to a proposed increase in the gas tax claimed that the increase would hurt working-class people the most since they commute the farthest to work. Suppose that the group randomly surveyed 24 individuals and asked them their daily one-way commuting mileage. The results are in idgh7333173. Using a 5 percent significance level, test the hypothesis that the three mean commuting mileages are the same.

Using a significance level of 5 percent, test the hypothesis that the four magazine types have the same mean length.

Eliminate one magazine type that you now feel has a mean length different from the others. Redo the hypothesis test, testing that the remaining three means are statistically the same. Use a new solution sheet. Based on this test, are the mean lengths for the remaining three magazines statistically the same?

1. H_a: µ_c = µ_n = µ_h
2. At least any two of the magazines have different mean lengths.
3. df(num) = 2, df(denom) = 12
4. F distribtuion
5. F = 15.28
6. p-value = 0.0005
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 mean lengths of the magazines are different.

A researcher wants to know if the mean times (in minutes) that people watch their favorite news station are the same. Suppose that eip-id2754369 shows the results of a study.

Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05.

Are the means for the final exams the same for all statistics class delivery types? eip-id1164473001430 shows the scores on final exams from several randomly selected classes that used the different delivery types.

Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05.

1. H₀: μ_o = μ_h = μ_f
2. At least two of the means are different.
3. df(n) = 2, df(d) = 13
4. F_2,13
5. 0.64
6. 0.5437
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: The mean scores of different class delivery are not different.

Are the mean number of times a month a person eats out the same for whites, blacks, Hispanics, and Asians? Suppose that fs-idm13269984 shows the results of a study.

Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05.

Are the mean numbers of daily visitors to a ski resort the same for the three types of snow conditions? Suppose that fs-idp109843184 shows the results of a study.

Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05.

1. H₀: μ_p = μ_m = μ_h
2. At least any two of the means are different.
3. df(n) = 2, df(d) = 12
4. F_2,12
5. 3.13
6. 0.0807
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: There is not sufficient evidence to conclude that the mean numbers of daily visitors are different.

Sanjay made identical paper airplanes out of three different weights of paper: light, medium, and heavy. He made four airplanes from each of the weights and launched them himself across the room. Here are the distances (in meters) that his planes flew.

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

Original figure description: the graph is a scatter plot which represents the data provided. The horizontal axis is labeled 'Distance in Meters,' and extends form 2 to 6. The vertical axis is labeled 'Weight of Paper' and has light, medium, and heavy categories.

1. Take a look at the data in the graph. Look at the spread of data for each group (light, medium, heavy). Does it seem reasonable to assume a normal distribution with the same variance for each group?
2. Why is this a balanced design?
3. Calculate the sample mean and sample standard deviation for each group.
4. Does the weight of the paper have an effect on how far the plane will travel? Use a 1 percent level of significance. Complete the test using the method shown in the bean plant example in element-349.
   • Variance of the group means __________
   • MS_between= ___________
   • Mean of the three sample variances ___________
   • MS_within = _____________
   • F statistic = ____________
   • df(num) = __________, df(denom) = ___________
   • Number of groups _______
   • Number of observations _______
   • p-value = __________ (P(F > _______) = __________)
   • Graph the p-value.
   • Decision: _______________________
   • Conclusion: _______________________________________________________________

DDT is a pesticide that has been banned from use in the United States and most other areas of the world. It is quite effective but persisted in the environment and over time proved to be harmful to higher-level organisms. Famously, egg shells of eagles and other raptors were believed to be thinner and prone to breakage in the nest because of ingestion of DDT in the food chain of the birds.

An experiment was conducted on the number of eggs (fecundity) laid by female fruit flies. There are three groups of flies. One group was bred to be resistant to DDT (the RS group). Another was bred to be especially susceptible to DDT (SS). The third group was a control line of nonselected or typical fruit flies (NS). Here are the data:

The values are the average number of eggs laid daily for each of 75 flies (25 in each group) over the first 14 days of their lives. Using a 1 percent level of significance, are the mean rates of egg selection for the three strains of fruit fly different? If so, in what way? Specifically, the researchers were interested in whether the selectively bred strains were different from the nonselected line, and whether the two selected lines were different from each other.

Here is a chart of the three groups:

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

Original figure description: This graph is a scatterplot which represents the data provided. The horizontal axis is labeled 'Mean eggs laid per day' and extends from 10 - 50. The vertical axis is labeled 'Fruitflies DDT resistant or susceptible, or not selected.' The vertical axis is labeled with the categories NS, RS, SS.

The data appear normally distributed from the chart and of similar spread. There do not appear to be any serious outliers, so we may proceed with our ANOVA calculations, to see if we have good evidence of a difference between the three groups.

H₀: μ₁ = μ₂ = μ₃

H_a: μ_i ≠ μ_j some i ≠ j

Define μ₁, μ₂, μ₃, as the population mean number of eggs laid by the three groups of fruitflies.

F statistic = 8.6657

p-value = 0.0004

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

Original figure description: This graph shows a nonsymmetrical F distribution curve. This curve does not have a peak, but slopes downward from a maximum value at (0, 1.0) and approaches the horiztonal axis at the right edge of the graph.

Decision: Since the p-value is less than the level of significance of 0.01, we reject the null hypothesis.

Conclusion: We have good evidence that the average number of eggs laid during the first 14 days of life for these three strains of fruitflies are different.

Interestingly, if you perform a two sample t test to compare the RS and NS groups they are significantly different (p = 0.0013). Similarly, SS and NS are significantly different (p = 0.0006). However, the two selected groups, RS and SS are not significantly different (p = 0.5176). Thus we appear to have good evidence that selection either for resistance or for susceptibility involves a reduced rate of egg production (for these specific strains) as compared to flies that were not selected for resistance or susceptibility to DDT. Here, genetic selection has apparently involved a loss of fecundity.

The data shown is the recorded body temperatures of 130 subjects as estimated from available histograms.

Traditionally, we are taught that the normal human body temperature is 98.6 °F. This is not quite correct for everyone. Are the mean temperatures among the four groups different?

Calculate 95 percent confidence intervals for the mean body temperature in each group and comment about the confidence intervals.