Question

1.. The average “psychological health” score for college professors in America is µ = 24 on a standardized psychological health survey. A statistics student from Valpo wants to know if professors from Valpo have psychological health scores that are significantly lower than the national average. The student has 36 Valpo professors complete the survey and finds that their mean to be 21 with a standard deviation of 5. Do a one-tailed test. Would you reject the null hypothesis in this study?

	a.	Yes
	b.	No

2. All CIs consist of two parts, namely, a ______ and a(n) ______.

	a.	point estimate; margin of error
	b.	point estimte; effect size
	c.	margin of error; p value
	d.	margin of error; effect size

3. Researchers obtained a sample of 36 college students who all have the same history instructor this semester. Half of the students were shown a 2-min video that claimed the purpose of education was to help students “learn how to learn” so that they can enjoy a lifetime of learning after college. The other half of students were shown a 2-min video that claimed the purpose of education was to teach facts to students. After watching the 2-min videos, the students were asked to rate their history instructor using a 10-point scale, 1 = very bad teacher to 10 = very good teacher. The mean and standard deviations for each group of 18 students are provided below. Use the provided information to answer the next four questions. Use an α of .05, two tailed.

“learn to learn” Group 1: M₁ = 5.9, SD₁ = 1.8, n₁ = 18

“learn facts” Group 2: M₂ = 7.2, SD₂ = 1.7, n₂ = 18

Which fact suggests that the homogeneity of variance assumption was met?

	a.	Students’ rating were not influenced by other students’ ratings.
	b.	The ratings of teachers are normally distributed.
	c.	The study clearly defined how the two groups were different and the ratings were measured on an interval/ratio scale.
	d.	The two sample SDs are similar.

Answer 1

(1) Null and Alternative Hypotheses 1.1) The following null and alternative hypotheses need to be tested: Ho: u = 24 Ha: 4 < 24 This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a left-tailed test is te = -1.69. The rejection region for this left-tailed test is R=t:t< -1.69 st Statistics The t-statistic is computed as follows: Ã – Mo 21 – 24 t = = s/V 57/58 = -3.6 (4) Decision about the null hypothesis Since it is observed that t = -3.6 <tc = -1.69, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0.0005, and since p= 0.0005 < 0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean u is less than 24, at the 0.05 significance level. Confidence Interval The 95% confidence interval is 19.308 < < 22.692.

Graphically. T-Test Results: t-stats = -3.6, p-value =0.0005 0.40 0.35 0.30 0.20 0.15 0.10 0.05 0.00 -4.0 -3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
1) Yes, we reject the null hypothesis.

2) All CI's consists of two parts, namely a point estimate and a(n) margin of error.

3) That fact suggest that-

C) The study clearly defined how the two groups were different and the ratings were measured on an interval / ratio scale.