# Testing Two Variances Worksheet

Testing Two Variances Worksheet
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1.
The basic assumption(s) for estimating the difference between two variances is/are:
I. The samples must be dependent on each other.
II. The samples must be independent of each other.
III. The populations from which the samples were obtained must be normally distributed.
IV. The populations from which the samples were obtained must depart from normality.
 a. I and III only b. II and III only c. II and IV only d. I and IV only

#### Solution:

The samples must be independent of each other.

The populations from which the samples were obtained must be normally distributed.

2.
A teacher claims that the variance of the final examination scores in her class is larger for male students than for the female students. The data from the final exam for the last semester reveals that the standard deviation of the scores for 30 males is 6.8 and that of 20 females is 4.6. Is there enough evidence to support her claim at 99% confidence level? Also give the F-value.
 a. support the claim, 2.84 b. support the claim, 2.185 c. reject the claim, 2.185 d. reject the claim, 2.84

#### Solution:

H0: σ12 ≤ σ22 and H1: σ12 > σ22
[Null and Alternative hypotheses.]

Standard deviation of scores of males (s1) = 6.8
Standard deviation of scores of females (s2) = 4.6
[Choose s1 and s2 such that s1 > s2.]

Number of males (n1) = 30
Number of females (n2) = 20

The test value F = s12s22 = 2.185
[Substitute the values and simplify.]

d.f.N = Degrees of freedom of numerator = n1 - 1 = 29
d.f.D = Degrees of freedom of denominator = n2 - 1 = 19

The critical value F(29, 19) at 99% confidence(α = 0.01) is 2.84
[This is one-tailed test.]

Since the test value is less than the critical value, we fail to reject the null hypothesis.

So, there is not enough evidence to support the claim that the variance of the scores of males is greater than that of the females.

3.
The data provided shows the number of people leaving two different companies in a year. Find whether the variance of the number of people leaving the companies differ at 90% and 80% confidence levels.

 a. do not differ, do not differ b. differ, do not differ c. do not differ, differ d. differ, differ

#### Solution:

H0: σ12 = σ22 and H1: σ12 ≠ σ22
[Null and Alternative hypotheses.]

Sample variance, s2 = Σ(x-X)2n-1, where X is the sample mean and n is the sample size.

Variance of employees leaving company 1(s12) = 0.386
Variance of employees leaving company 2 (s22) = 0.333
[Use the formula for finding the variances.]

Number of months in a year (n1) = 12
Number of months in a year (n2) = 12

d.f.N = n1 - 1 = 12 - 1 = 11
d.f.D = n2 - 1 = 12 - 1 = 11

Critical value F(11, 11) at 90% confidence, α = 0.05 is 2.79
Critical value F(11, 11) at 80% confidence, α = 0.1 is 2.21
[Since it is a two tailed test, consider α2.]

The test value F = s12s22 = 1.159
[Substitute and simplify.]

Since the test value is less than the critical value for both 90% and 80% confidence levels, we fail to reject the null hypothesis.

So, there is not enough evidence to say that the two variances differ significantly.

4.
Find the F-test value and the critical value at 95% confidence for testing the population variances. The data relating to samples are shown.
 Sample size Sample variance 30 12 25 8

 a. 1.2, 2.15 b. 2.15, 1.2 c. 1.94, 1.5 d. 1.5, 1.94

#### Solution:

s12 = 12, s22 = 8, n1 = 30, n2 = 25

d.f.N = n1 - 1 = 29
d.f.D = n2 - 1 = 24

At 95% confidence, α = 0.05, the critical value is F(29, 24) = 1.94
[Refer the F distribution table.]

The test value F = s12s22 = 1.5
[Substitute the values and simplify.]

So, the test value is 1.5 and the critical value is 1.94.

5.
If the degrees of freedom of the numerator is 15 and that of the denominator is 24 for testing the variances of two populations using F-test, then find the test value and the critical value at 95% confidence. The sample variances are 45 and 78.
 a. 2.65, 1.83 b. 1.73, 2.11 c. 2.11 1.73 d. 1.83, 2.65

#### Solution:

s12 = 78, s22 = 45, d.f.N = 15, d.f.D = 24
[Choose s1 and s2 such that s1 > s2.]

The critical value for 95% confidence, α = 0.05 is F(15, 24) = 2.11
[Refer to the F distribution table.]

The test value F = s12s22 = 1.73
[Substitute and simplify.]

So, the test value is 1.73 and the critical value is 2.11.

6.
Which of the following statement(s) is/are true?
I. $z$ test is used for testing two means and two variances.
II. F test is used for testing two variances.
III. $t$ test is used for testing two variances.
 a. I only b. all are correct c. I and II only d. II only

#### Solution:

F test is used for the comparison of two variances or standard deviations.

z or t tests are used for comparing two means.

7.
Which of the following is/are the characteristics of F distribution?
I. It is negatively skewed.
II. The values are always positive or zero.
III. The mean value of F is approximately equal to one.
 a. I only b. II and III only c. I and II d. II only

#### Solution:

The values of F cannot be negative because variances are always either positive or zero.

The distribution is positively skewed.

The mean value of F is approximately equal to 1.

The F distribution is a family of curves based on the degrees of freedom of the variance of the numerator and the degrees of freedom of the variance of the denominator.

So, statements II and III are correct.

8.
Which of the following is/are true?
I. The mean value of F is approximately zero.
II. F distribution is a family of curves based on the degrees of freedom of the variance of the numerator.
III. F distribution is a family of curves based on the degrees of freedom of the variance of the denominator.
 a. I and II only b. II and III only c. I only d. neither I nor II

#### Solution:

The mean value of F is approximately 1.

The F distribution is a family of curves based on the degrees of freedom of variance of the numerator and the degrees of freedom of the variance of the denominator.

9.
A manufacturer claims that the variance of defective detergent packets in cartons produced by two companies differ significantly. A preliminary investigation revealed that the sample variance of the number of defective packets in 20 cartons of company 1 as 69 and that of 25 cartons of company 2 as 75. Check the claim at 90% and 95% confidence levels.
 a. do not differ, do not differ b. differ, do not differ c. do not differ, differ d. differ, differ

#### Solution:

H0: σ12 = σ22 and H1: σ12 ≠ σ22
[Null and Alternative hypotheses.]

Variance of company 2 is s12 = 75
Variance of company 1 is s22 = 69
[Choose s1 and s2 such that s1 > s2.]

Number of packets manufactured by company with largest variance n1 = 25
Number of packets manufactured by other company n2 = 20

The test value F = s12s22 = 1.09
[Substitute the values and simplify.]

d.f.N = Degrees of freedom of numerator = n1 - 1 = 24
d.f.D = Degrees of freedom of denomiator = n2 - 1 = 19

The critical value F(24, 19) at 90% confidence, α = 0.05 is 2.11
The critical value F(24, 19) at 95% confidence, α = 0.025 is 2.45
[Since this is two tailed, we consider α2.]

Since the test value is less than the critical values for both α = 0.05 and α = 0.025, we fail to reject the null hypothesis.

So, there is not enough evidence to say that the variances of defective packets of the products produced by both the companies differ significantly at 90% and 95% confidence levels.

10.
A statistician wishes to see whether the variance of the number of passengers traveling from a particular city to city 1 is different from that to city 2. On observation, it is found that the variance of the number of passengers traveling to city 1 is 250.6 for 30 days and that to city 2 is 105.3 for 34 days. Do the variances differ at 98% confidence interval? Give the F-value also.
 a. differ significantly, 2.33 b. do not differ significantly, 2.33 c. do not differ significantly, 2.38 d. differ significantly, 2.38

#### Solution:

H0: σ12 = σ22 and H1: σ12 ≠ σ22
[Null and Alternative hypotheses.]

n1 = number of days observed for city 1 = 30
n2 = number of days observed for city 2 = 34

s12 = sample variance of passengers to city 1 = 250.6
s22 = sample variance of passengers to city 2 = 105.3

The test value F = s12s22 = 2.38
[Substitute and simplify.]

d.f.N = degrees of freedom of numerator = n1 - 1 = 29
d.f.D = degrees of freedom of denominator = n2 - 1 = 33

The critical value F(29,33) = 2.333 at 98% confidence, α = 0.01
[Since it is two tailed test consider α2.]

The test value is greater than the critical value and therefore we reject the null hypothesis.

So, there is enough evidence to say that the variances of passengers travelling to both the cities differ significantly at a confidence level of 98%.