Question

A group of researchers found that there were lower ratings of faculty who taught online courses across the US from students. To verify this, the researchers collected a sample of n = 25 student evaluations during the Fall 2020 semester and obtained an average rating of M = 7.29. The complete set of reviews during the previous year averaged µ = 7.52 with a standard deviation of σ = 0.60. Can the researchers conclude that student evaluations of faculty during online courses are lower than the general population average (reject the null)? Use a one-tailed test with α = .05.

a) Yes
b) No
c) Not enough information to determine.

Answer 1

To determine if there is a significant difference between student evaluations of faculty during online courses and the general population average, the researchers need to conduct a hypothesis test. The test statistic (-1.33) is greater than the critical value (-1.645), so we fail to reject the null hypothesis. Therefore, we do not have enough evidence to conclude that student evaluations of faculty during online courses are lower than the general population average.

Step-by-step explanation:

To determine if there is a significant difference between student evaluations of faculty during online courses and the general population average, the researchers need to conduct a hypothesis test.

1. The null hypothesis (H0) states that there is no difference between the mean rating of online courses and the general population average. The alternative hypothesis (Ha) states that the mean rating of online courses is lower than the general population average.
2. Since the population standard deviation is known (σ = 0.60) and the sample size (n = 25) is small, a z-test can be used to perform the hypothesis test.
3. To calculate the test statistic, use the formula: z = (M - µ) / (σ / sqrt(n)), where M is the sample mean, µ is the population mean, σ is the population standard deviation, and n is the sample size.
4. With the given data, the test statistic is z = (7.29 - 7.52) / (0.60 / sqrt(25)) = -1.33.
5. Finally, compare the test statistic to the critical value from the z-table at the desired significance level (α = 0.05) to make a decision. In this case, the critical value for a one-tailed test is -1.645.

The test statistic (-1.33) is greater than the critical value (-1.645), so we fail to reject the null hypothesis. Therefore, we do not have enough evidence to conclude that student evaluations of faculty during online courses are lower than the general population average.