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?
With infinite degrees of freedom, you need a t score of 1.96 to reject the null hypothesis at the .05 level. What does that tell you, if your t = -.20?
Reject it.
To make a decision regarding the null hypothesis, we need to perform a t-test to determine the statistical significance of the results.
Given:
- The null hypothesis: Fourth graders in this class are less depressed than those at the school generally.
- Sample size (n): 20 students
- Sample mean (x̄): 4.4
- Population mean (μ): 5
- t-score: -0.20
We can calculate the t-value using the formula:
t = (x̄ - μ) / (s / sqrt(n))
where s is the sample standard deviation.
However, the variation is not known in this case, so it is not possible to calculate the t-value accurately.
Without the sample standard deviation, we cannot determine the p-value associated with the t-score and compare it with an alpha level (e.g., 0.05) to make a decision regarding the null hypothesis.
Therefore, in this situation, we cannot make a decision regarding the null hypothesis based on the information provided.
To determine the decision regarding the null hypothesis, the counselor needs to conduct a hypothesis test using the t-test statistic.
The null hypothesis (H0) in this case is that the mean depression score of fourth graders in this class is equal to the mean depression score of fourth graders in general at her school.
The alternative hypothesis (Ha) would be that the mean depression score of fourth graders in this class is less than the mean depression score of fourth graders in general at her school.
Given that the sample size is 20 and the mean depression score of the sample is 4.4, the t-score is calculated as follows:
t = (sample mean - population mean) / (sample standard deviation / square root of sample size)
But the variation of the fourth graders in general at her school is not known, so the sample standard deviation cannot be calculated. Therefore, it is not possible to calculate the t-score directly.
However, the counselor has already calculated a t-score of -0.20. It is important to note that the t-score is negative, indicating that the sample mean of 4.4 is lower than the population mean of 5 (as mentioned in the question).
Now, the counselor needs to compare the calculated t-score (-0.20) with the critical value from the t-distribution table. The critical value is determined based on the desired level of significance (alpha) and the degrees of freedom (which is n - 1 in this case, where n is the sample size).
Suppose the counselor chooses a significance level of alpha = 0.05. From the t-distribution table with degrees of freedom (df) = 19 (20 - 1), the critical value for a one-tailed test is approximately -1.729 (assuming a left-tailed test since Ha suggests the sample mean is less than the population mean).
Since the calculated t-score (-0.20) is not less than the critical value (-1.729), the counselor fails to reject the null hypothesis, meaning there is not enough evidence to support that fourth graders in this class are less depressed than fourth graders in general at her school.