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. | 31.410 | ||

c. | 10.117 | ||

d. | 10.851 |

[Use the chi-square distribution table.]

Correct answer : (2)

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. | 32.000 | ||

d. | 30.575 |

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

The chi-square critical value for a left tailed test with

Correct answer : (1)

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 |

[

The chi-square critical values for a two tailed test with

Correct answer : (4)

4.

What is the critical value of the chi-square test, H_{o} : s^{2} > ≤ 100 and H_{1} : s^{2} > 100 for $n$ = 15 at α = 0.025?

a. | 5.629 | ||

b. | 26.119 | ||

c. | 6.262 | ||

d. | 27.488 |

[Null and Alternate hypotheses.]

So, the test is right tailed.

The critical value of the chi-square for a right tailed test with

Correct answer : (2)

5.

What is the critical value of the chi-square test, H_{0} : s^{2} ≥ 81 and H_{1}: s^{2} < 81 for $n$ = 26 at α = 0.05?

a. | 14.611 | ||

b. | 37.652 | ||

c. | 38.885 | ||

d. | 15.379 |

[Null and Alternate hypotheses.]

So, the test is left tailed.

The critical value of the chi-square for a left tailed test with

Correct answer : (1)

6.

In a statistical hypothesis testing for a variance, the test value of chi-square, c^{2} = 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 |

c

[Use the chi-square distribution table.]

The corresponding

Therefore, 0.01 <

Correct answer : (1)

7.

In a statistical hypothesis testing for a variance, the test value of chi-square, c^{2} = 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 |

c

The corresponding

Since the test is left tailed,

1 - 0.99 = 0.01 and 1 - 0.975 = 0.025

Therefore, 0.01 <

Correct answer : (2)

8.

In a statistical hypothesis testing for a variance , the test value of chi-square, c^{2} = 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 |

c

[Use the chi-square distribution value.]

The corresponding

Therefore, 0.01 <

Correct answer : (3)

9.

In a statistical hypothesis test, $s$^{2} is the sample variance, s^{2} 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. | $\frac{(n-1){s}^{2}}{{\sigma}^{2}}$ | ||

c. | $\frac{(n-1){\sigma}^{2}}{{s}^{2}}$ | ||

d. | $\frac{(n+1){s}^{2}}{{\sigma}^{2}}$ |

Correct answer : (2)

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 |

[Null and alternate hypotheses.]

Test value, c

c

[Substitute the values and simplify.]

Degrees of freedom =

c

Therefore, 0.05 <

[Use calculator.]

The

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.

Correct answer : (1)