In Fall 2016, I decided to change up my teaching strategy a bit. I changed from
ID: 3060358 • Letter: I
Question
In Fall 2016, I decided to change up my teaching strategy a bit. I changed from a purely lecture-based course to a more active learning environment. In the past, the final exam mean test score across all my courses (the population) was an 85, with a standard deviation of 9. At the end of the Fall 2016 semester, my new active learning class had a mean final exam score of 89 with a standard deviation of 6 (n = 36). Calculate the appropriate test statistic for my new active learning class. Only include the test statistic's numerical value and carry out your rounding to the hundredths place after the decimal. For example, 9.25. You only have two test statistic options: a t-test or z-test
Based on your test statistic calculation, did my active learning class earn a significantly higher final exam score as compared to the population? Why or why not? In your answer, be sure to mention what the critical value was and whether it was a one-tailed or two-tailed test
Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: < 85
Alternative hypothesis: > 85
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample z-test.
Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).
SE = s / sqrt(n)
S.E = 1.00
z = (x - ) / SE
z = 4.0
zcritical = 1.645
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
The observed sample mean produced a z statistic test statistic of 4.00.
Interpret results. Since the z-value (4.00) is greater than the z critical(1.645), we have to reject the null hypothesis.
From the above test we have sufficient evidence in the favor of the claim that active learning class earn a significantly higher final exam score as compared to the population.