4. Suppose that previous studies have found that the mean amount of sessions necessary to “cure” depressive patients using a particular therapeutic model is 25 sessions, with a standard deviation (σ) of 4. A new model is introduced, and you want to determine whether the amount of sessions needed to “cure” depressive patients with this new model is significantly lower than 25. You select 40 patients, all of whom receive therapy using the new model. You keep track of the amount of sessions these participants take to become “cured”. The test statistic was -3.90.
If you find that a situation requires testing the null hypothesis using the Normal Curve, you don't need to figure degrees of freedom, so you don't have to do (c) for that problem(s). Also, if you find that a problem(s) require(s) testing the null hypothesis using ANOVA, then you don't have to state whether the test is bilateral or unilateral.
Finally, assume that any information that is not provided is therefore unknown to the researcher. Also, assume that the level of significance (alpha) used by the researcher is 0.05.
For each of the following situations, determine:
a) the null and research hypotheses;
b) the hypothesis test used (normal curve, single sample t-test, dependent samples t-test, independent samples t-test, ANOVA). Also state whether the test is unilateral or bilateral;
c) the degrees of freedom
d) the critical value (cutoff score on the comparison distribution);
e) your sample’s test score;
f) your conclusion on whether to accept or reject the null hypothesis (you must show how the comparison of your cutoff score with your sample’s test score leads to your conclusion).
From your first paragraph, I don't know what test statistics you are using. If Z = -3.9, then it is definitely significant at P = .05.
I don't know what the "following situations" are.
Sorry that I can't help more.
a) The null hypothesis (H0): The mean amount of sessions needed to "cure" depressive patients using the new therapeutic model is not significantly lower than 25.
The research hypothesis (Ha): The mean amount of sessions needed to "cure" depressive patients using the new therapeutic model is significantly lower than 25.
b) The hypothesis test used: Single sample t-test
The test is unilateral because we are testing whether the new therapeutic model is significantly lower than 25 sessions.
c) The degrees of freedom: For a single sample t-test, the degrees of freedom (df) is calculated as (n-1), where n is the number of observations in the sample. In this case, n is 40, so the degrees of freedom are 39.
d) The critical value: The critical value is the cutoff score on the comparison distribution, which determines the boundary for rejecting the null hypothesis. Since the level of significance (alpha) is 0.05, we need to find the critical t-value from a t-table or t-distribution with 39 degrees of freedom. Looking up the critical value for a one-tailed test at alpha=0.05, we find the critical t-value to be approximately -1.686.
e) The sample's test score: The test statistic given in the question is -3.90.
f) Conclusion: To determine whether to accept or reject the null hypothesis, we compare the sample's test score (-3.90) with the critical value (-1.686) for a one-tailed test. Since the test score is beyond the critical value, we reject the null hypothesis. This means that there is sufficient evidence to conclude that the mean amount of sessions needed to "cure" depressive patients using the new therapeutic model is significantly lower than 25.