Part 1: Comparing the average number of polling quizzes answered by class attendees vs. live-streamers

Answer 1: Which test is correct to use, if we know the population standard deviation?

If we know the population standard deviation, the correct test to use in this scenario is a two-sample Z-test. This is because we are comparing the means of two independent groups (students attending classes vs. live-streamers) and have information about the population standard deviation.

Answer 2: What is the null hypothesis?

The null hypothesis for this test would be: $H_0: \mu_{\text{class}} = \mu_{\text{live-stream}}$ This means that the average number of polling quizzes answered by students attending classes is equal to the average number of polling quizzes answered by live-streamers.

Answer 3: The p-value of the test is 0.24. What is the decision of the test?

Given a p-value of 0.24 and assuming a typical significance level of $\alpha = 0.05$, we fail to reject the null hypothesis. This means there is no statistically significant evidence to suggest that there is a difference in the average number of polling quizzes answered between students attending classes and live-streamers.

Part 2: Testing if the average number of polling quizzes answered is equal to 38

Answer 4: Which test is correct to use, if we do not know the population standard deviation?

If we do not know the population standard deviation, the correct test to use is a one-sample t-test. This is because we are testing whether the sample mean differs from a hypothesized population mean (38), and the population standard deviation is unknown.

Answer 5: The test statistic is -2.67. This means that the difference between the sample mean and the possible population mean, as stated in the null hypothesis (i.e. 38), is:

The test statistic of -2.67 means that the sample mean is 2.67 standard errors below the hypothesized population mean of 38. In other words, the sample mean is significantly lower than 38, based on the number of standard errors.

Answer 6: The p-value of the test is 0.008. What is the decision of the test?

Since the p-value of 0.008 is less than the typical significance level of $\alpha = 0.05$, we reject the null hypothesis. This means that there is statistically significant evidence to suggest that the average number of polling quizzes answered is not equal to 38, and the sample mean is likely lower than 38.

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Do you want any more details or have any questions?

Here are five related questions that expand on this:

1. What assumptions must be verified before conducting a two-sample Z-test?
2. How do you interpret a confidence interval in the context of hypothesis testing?
3. What are the consequences of using a Z-test instead of a t-test when the population standard deviation is unknown?
4. How does sample size affect the validity of the Z-test and t-test results?
5. What would be the decision if the p-value were 0.06 in Part 2?

Tip: When conducting hypothesis tests, always ensure that the assumptions (e.g., normality, independence, and known/unknown population standard deviation) are met to avoid incorrect conclusions.