Question

A national report indicates that the mean and standard deviation of the ACT scores of incoming freshman at American public universities are 24.6 and 6.2, respectively. A sample of 41 incoming freshmen at a small, suburban college provide a sample mean of 22.8 and a sample standard deviation of 7.7. The researcher is interested in finding evidence at the .05 level that the standard deviation in ACT scores is greater than the reported value for all public universities.

. What is the hypothesis, test statistic, pvalue, critical value, and decision?

Answer 1

Let be the true standard deviation of the ACT scores for the suburban college.

The researcher is interested in finding evidence at the .05 level that the standard deviation in ACT scores is greater than the reported value for all public universities (which is 6.2)

The hypotheses are

We have the following information from the sample

s=7.7 is the sample standard deviation for the suburban college

n=41 is the sample size

is the hypothesized value of standard deviation for the suburban college

The chi-square test statistics is

The degrees of freedom for the chi-square statistics is n-1=41-1=40.

This is a right tailed (one tailed) test (The alternative hypothesis has ">"). Hence the critical value for alpha=0.05 is the area under the right tail of a chi-square distribution.

From the chi-square table with df=40 and alpha=0.05 we get as the critical value.

That means we reject the null hypothesis, if the test statistics value is greater than 55.7585. Here since the test statistics of 61.70 is greater than 55.7585, we reject the null hypothesis, while using the critical value method.

We also look at the p-value. This is a right tailed test. The p-value is

Using the chi-square table we can approximate for df=40.  We find that

That means the true value of is between 0.01 and 0.025.

p-value is between 0.01-0.025.

To find the exact p-value we can use excel function =CHISQ.DIST.RT(61.7,40) and get p-value=0.015

We will reject the null hypothesis if the p-value is less than the significance level of alpha=0.05. Here the p-value is less than 0.05 and hence we reject the null hypothesis.

Using both the critical value method and p-value method, we reject the null hypothesis.

We conclude that there is sufficient evidence to support the claim that standard deviation of ACT scores for suburban college is greater than the reported value for all public universities.