A school counselor tests the level of depression in fourth graders in a particular class of 20 students. The counselor wants to know whether the kind of students in this class differs from that of fourth graders in general at her school. On the test, a score of 10 indicates severe depression, while a score of 0 indicates no depression. From reports, she is able to find out about past testing. Fourth graders at her school usually score 5 on the scale, but the variation is not known. Her sample of 20 fiftsh graders has a mean depression score of 4.4.

Question

A school counselor tests the level of depression in fourth graders in a particular class of 20 students. The counselor wants to know whether the kind of students in this class differs from that of fourth graders in general at her school. On the test, a score of 10 indicates severe depression, while a score of 0 indicates no depression. From reports, she is able to find out about past testing. Fourth graders at her school usually score 5 on the scale, but the variation is not known. Her sample of 20 fiftsh graders has a mean depression score of 4.4.

Suppose the counselor tested the null hypothesis that fourth graders in this class were less depressed than those at the school generally. She figures her t scores to be -.20. What decision should she make regarding the null hypothesis

Answer 1

If she has the mean for fourth graders, why does she get a sample of fifth graders? How does she get a t value without any indication of variation from the sample?

To be significant at P = .05, t with 19 or (n-1) df, you would need a value of 2.093.

What would you conclude?

Answer 2

To make a decision regarding the null hypothesis, the counselor can perform a hypothesis test using the t-distribution.

First, let's state the null and alternative hypotheses:
Null hypothesis (H0): The mean depression score for fourth graders in this class is equal to the mean depression score for fourth graders at the school generally.
Alternative hypothesis (Ha): The mean depression score for fourth graders in this class is less than the mean depression score for fourth graders at the school generally.

Next, we need to determine the level of significance or alpha (α) to use. This value determines how confident we need to be in order to reject the null hypothesis. Common choices for alpha are 0.05 (5%) or 0.01 (1%).

Now let's calculate the t-statistic. The formula for the t-statistic is:

t = (sample mean - population mean) / (sample standard deviation / √sample size)

Given that the sample mean depression score is 4.4, the population mean is 5 (as mentioned in the problem statement), and the t-score is -0.20, we need the sample standard deviation to calculate the t-statistic.

The problem doesn't provide information about the sample standard deviation, so we cannot proceed with the hypothesis test based on the given information.

However, if the sample standard deviation were provided, the counselor could calculate the t-statistic and compare it to the critical t-value corresponding to the chosen alpha level and degrees of freedom.

If the t-statistic falls within the critical region (t < critical t-value), then we reject the null hypothesis in favor of the alternative hypothesis. If it falls outside the critical region, we fail to reject the null hypothesis.