Exhibit 11-3
The contents of a sample of 26 cans of apple juice showed a standard deviation of 0.06 ounces. We are interested in testing to determine whether the variance of the population is significantly more than 0.003.
-Refer to Exhibit 11-3. At 95% confidence, the null hypothesis
A)should be rejected
B)should not be rejected
C)should be revised
D)None of these alternatives is correct.

Question

Quizplus · Accepted Answer

At 95% confidence, we would use a significance level of 0.05 for a two-tailed test. The null hypothesis is that the population variance is equal to 0.003, so we would use a chi-squared test with 25 degrees of freedom (n-1) to test this hypothesis. The critical values for a chi-squared distribution with 25 degrees of freedom at a significance level of 0.025 (divided by 2 for a two-tailed test) are 11.070 and 39.364. The test statistic can be calculated as (n-1) * sample variance / population variance, which gives us (26-1) * 0.06^2 / 0.003 = 67.2. Since 67.2 is greater than 39.364, we fail to reject the null hypothesis and conclude that there is not enough evidence to support the claim that the population variance is significantly more than 0.003. Therefore, the answer is B, the null hypothesis should not be rejected.

Exhibit 11-3 The Contents of a Sample of 26 Cans of Apple