﻿ Variance Standard Deviation Worksheet | Problems & Solutions

# Variance Standard Deviation Worksheet

Variance Standard Deviation Worksheet
• Page 1
1.
Find the critical value of the chi-square test for a variance with 20 degrees of freedom, at α = 0.05 and test is right tailed.
 a. 30.144 b. 31.41 c. 10.117 d. 10.851

#### Solution:

The chi-square critical value for a right tailed test with df = 20 and α = 0.05 is 31.410
[Use the chi-square distribution table.]

2.
Find the critical value of the chi-square test for a variance with 16 degrees of freedom, at α = 0.01 and the test is left tailed.
 a. 5.812 b. 5.229 c. 32 d. 30.575

#### Solution:

1 - α = 1 - 0.01 = 0.99
[Find (1 - α) value for a left tailed test.]

The chi-square critical value for a left tailed test with df = 16 at 0.99 is 5.812

3.
Find the critical values of the chi-square test for a variance with 28 degrees of freedom, α = 0.10 and the test is two tailed.
 a. 41.337, - 41.337 b. 18.939, 37.916 c. 18.114, 36.741 d. 16.928, 41.337

#### Solution:

12 α = 12 (0.10) = 0.05 and (1 - 12 α) = 1 - 0.05 = 0.95.
[Find α2 and (1 - α2) for a two tailed test.]

The chi-square critical values for a two tailed test with df = 28 at 0.95 is 16.928 and at 0.05 is 41.337.

4.
What is the critical value of the chi-square test, Ho : s2 > ≤ 100 and H1 : s2 > 100 for $n$ = 15 at α = 0.025?
 a. 5.629 b. 26.119 c. 6.262 d. 27.488

#### Solution:

H0 : s2 ≤100 and H1 : s2 > 100
[Null and Alternate hypotheses.]

So, the test is right tailed.

df = n - 1 = 15 - 1 = 14 and α = 0.025

The critical value of the chi-square for a right tailed test with df = 14 at α = 0.025 is 26.119.

5.
What is the critical value of the chi-square test, H0 : s2 ≥ 81 and H1: s2 < 81 for $n$ = 26 at α = 0.05?
 a. 14.611 b. 37.652 c. 38.885 d. 15.379

#### Solution:

H0 : s2 ≥ 81 and H1 : s2 < 81
[Null and Alternate hypotheses.]

So, the test is left tailed.

df = n - 1 = 26 - 1 = 25 and α = 0.05, 1 - α = 1 - 0.05 = 0.95

The critical value of the chi-square for a left tailed test with df = 25 at 0.95 is 14.611.

6.
In a statistical hypothesis testing for a variance, the test value of chi-square, c2 = 32.426, $n$ = 18, and the test is right tailed. Find the P-value interval for the test.
 a. 0.01 < $p$-value < 0.025 b. 0.02 < $p$-value < 0.05 c. information is not sufficient d. 0.975 < $p$-value < 0.99

#### Solution:

The test value, c2 = 32.426, df = n - 1 = 18 - 1 = 17

c2 value falls between 30.191 and 33.409.
[Use the chi-square distribution table.]

The corresponding α values are 0.025 and 0.01.

Therefore, 0.01 < p-value < 0.025

7.
In a statistical hypothesis testing for a variance, the test value of chi-square, c2 = 9.428, degrees of freedom 20, and the test is left tailed. Find the $p$-value interval for the test.
 a. 0.975 < $p$-value < 0.99 b. 0.01 < $p$-value < 0.025 c. 0.02 < $p$-value < 0.05 d. information is not sufficient

#### Solution:

The test value, c2 = 9.428, df = 20

c2 value falls between 8.260 and 9.591

The corresponding α values are 0.99 and 0.975.

Since the test is left tailed, subtract the corresponding α values from 1 to get the p-value interval.

1 - 0.99 = 0.01 and 1 - 0.975 = 0.025

Therefore, 0.01 < p-value < 0.025

8.
In a statistical hypothesis testing for a variance , the test value of chi-square, c2 = 47.831, $n$ = 28 and the test is two tailed. Find the $P$-value interval for the test.
 a. information is not sufficient b. 0.005 < $p$-value < 0.01 c. 0.01 < $p$-value < 0.02 d. 0.990 < $p$-value < 0.995

#### Solution:

The test value, c2 = 47.831, df = n - 1 = 28 - 1 = 27

c2 value falls between 46.963 and 49.645
[Use the chi-square distribution value.]

The corresponding α values are 0.01 and 0.005

Since the test is two tailed, multiply both the α values with 2.

Therefore, 0.01 < p-value < 0.02

9.
In a statistical hypothesis test, $s$2 is the sample variance, s2 the population variance and $n$ the sample size, than the formula for the chi-square test for a single variance is:
 a. $\frac{n{s}^{2}}{{\sigma }^{2}}$ b. $\frac{\left(n-1\right){s}^{2}}{{\sigma }^{2}}$ c. $\frac{\left(n-1\right){\sigma }^{2}}{{s}^{2}}$ d. $\frac{\left(n+1\right){s}^{2}}{{\sigma }^{2}}$

#### Solution:

Formula for chi-square test for a single variance is c2 = (n-1)s2σ2, where s2 is the sample variance, σ2 the population variance and n the sample size.

10.
A survey on hospitals concluded that the standard deviation of the amount of time a patient has to wait before being attended by emergency room personnel is more than 5.4 minutes. Waiting times in the case of a sample of 16 patients are recorded and it was found that the standard deviation was 6.6 minutes. Find the P-value for the test. At α = 0.05, is there enough evidence to support the hospital survey report? (Assume that the variable is normally distributed.)
 a. 0.0976 ; no b. 0.9024 ; yes c. 0.9024 ; no d. 0.0976 ; yes

#### Solution:

H0 : s ≤ 5.4 and H1 : s > 5.4
[Null and alternate hypotheses.]

Test value, c2 = (n-1)s2σ2, where s2 is the sample variance, s2 the population variance and n the sample size.

n = 16, s = 5.4 and s = 6.6

c2 = (16-1)(6.6)2(5.4)2 = 22.407
[Substitute the values and simplify.]

Degrees of freedom = n - 1 = 16 - 1 = 15

c2 value falls between 22.307 and 24.996 and the corresponding α values are 0.10 and 0.05

Therefore, 0.05 < p-value < 0.10

p-value = 0.0976
[Use calculator.]

The p-value 0.0976 > 0.05, do not reject the null hypothesis.

So, there is not enough evidence to support the hospital survey report that the standard deviation of waiting times of patients is more than 5.4 minutes.
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