Question

A golfer would like to test the hypothesis that the variance of his golf score equals 10.0. A random sample of 28 rounds of golf had a sample standard deviation of 3.5. Using α = 0.10, the conclusion for this hypothesis test would be that because the test statistic is ________________. between the lower and upper critical values, we reject the null hypothesis and cannot conclude that the variance for this golfer's scores is not equal to 10.0 more than the upper critical value, we can reject the null hypothesis and conclude that the variance for this golfer's scores is not equal to 10.0 less than the lower critical value, we fail to reject the null hypothesis and conclude that the variance for this golfer's scores is not equal to 10.0 between the lower and upper critical values, we fail to reject the null hypothesis and cannot conclude that the variance for this golfer's scores is not equal to 10.0 Using α = 0.10, the lower critical value for this hypothesis test would be ________. 17.334 16.151 13.848 12.401 The test statistic for this hypothesis test would be ________. 33.08 40.75 12.57 46.23 The correct hypothesis statement would be ______. H0: σ2 = 10.0; H1: σ2 ≠ 10.0 H0: μ = 10.0; H1: μ ≠ 10.0 H0: σ = 10.0; H1: σ ≠ 10.0 H0: σ2 ≤ 10.0; H1: σ2 > 10.0

Answer 1

Answer:

Step-by-step explanation:

Hello!

Interest hypothesizes is that the variance of the golfer's scores equals to δ²= 10.0.

A random sample of 28 rounds of golf had a sample standard deviation S= 3.5

The statistics hypotheses are:

H₀: δ²= 10.0

H₁: δ²≠ 10.0

α: 0.05

To conduct a hypothesis test for the population variance you have to work using the Chi-Square distribution, this test is two-tailed so you will have two critical values. Between these two values is defined as the "not rejection region" and below and above them lies the "rejection region"

Lower critical value: $X^2_{n-1; \alpha /2}= X^2_{27; 0.05}= 16.928$

Upper critical value: $X^2_{n-1;1-\alpha /2}= X^2_{27;0.95}= 40.113$

If $X^2_{H_0}$ ≤ 16.928 or $X^2_{H_0}$ ≥ 40.113, the decision is to reject the null hypothesis.

If 16.928 < $X^2_{H_0}$ < 40.113, the decision is to not reject the null hypothesis.

$X^2= \frac{(n-1)S^2}{Sigma^2} ~~X^2_{n-1}$

$X^2_{H_0}= \frac{(28-1)12.25}{10} = 33.075$

The calculated $X^2_{H_0}$ is between the two critical values, so the decision is to not reject the null hypothesis. We can conclude that the population variance fro this golfers play score is 10.

I hope this helps!