Question

In a school district, all sixth grade students take the same standardized test. The superintendant of the school district takes a random sample of 29 scores from all of the students who took the test. She sees that the mean score is 167 with a standard deviation of 11.4238. The superintendant wants to know if the standard deviation has changed this year. Previously, the population standard deviation was 18. Is there evidence that the standard deviation of test scores has decreased at the α=0.05 level? Assume the population is normally distributed. Step 1 of 5: State the null and alternative hypotheses. Step 2 of 5: Determine the critical value(s) of the test statistic. If the test is two-tailed, separate the values with a comma. Round your answer to three decimal places. Step 3 of 5: Determine the value of the test statistic. Round your answer to three decimal places. Step 4 of 5: Make the decision. Step 5 of 5: What is the conclusion?

Answer 1

We have the following data given

n=29 represent the sample size selected for the test

α=0.05 represent the confidence level

s^2= 11.4238^2= 130.5032 represent the sample variance obtained

represent the value that we want to test

System of hypothesis

We want to review if the population deviation is significantly lower than 18 and we can check this with the variance, the system of hypothesis would be:

Null Hypothesis:

Alternative hypothesis:

Statistic

To test this hypothesis the statistic is:

The degrees of freedom for this case are:

df=n-1=29-1=28

Replacing in the statistic formula we got:

=11.2780

P value

since we have a left tailed test the p value would be givne by:

We can find the p value with this excel code:

=CHISQ.DIST(11.278,28,TRUE)

Since the p value is very low compared with the significance level provided 0.05, we have enough evidence to conclude that the true variance and on this case the deviation is significantly lower than 18 at 5% of significance.