hypothesis testing with one sample t-tests

Question

Someone is testing his new method of teaching and wants to see if it is effective. In the last class that he gave a particular exam, he did not use this new teaching method and that class got an average of 40 on the exam. After using the new method, he looks at 6 of his exam scores and gets the following results: 42, 47, 49, 47, 45, 46.

a. Identify the two groups:

The two groups are the old teaching method and the new teaching method.

What is the dependent variable?

The dependent variable is the

b. State the null and research hypothesis:

Null: The students who ____________________ will have _____________ the students who _________________

Research: The students who ____________________ will have _____________ the students who _________________

c. Figure out the numbers you need:

What is the sample size (N)?______________________

What is the POPULATION mean (μ)? ________

What is the SAMPLE mean (M)? (you need to figure this out on your own) ___________

Now, what is the sum of squares? (X – M)² ____________________________

What is the degrees of freedom? (N-1) _________

What is the variance? (X – M)² / (N-1) _________________________

What is the standard deviation? s = (X – M)² / (N-1)____________________

What is the standard error? s_M = s/ (hint: use the standard deviation you just found) __________

Now, you have everything you need for

d. calculate the t-statistic (t = (M – μ)/s_M) __________________________

e. What is the critical value? Use the value you got from N-1, and look for the number that is under .05 ___________________________

f. Compare your t value to the critical value. If the t is greater than the critical value, you __________. If the t is lower than the critical value, you _____________

What decision do you make here? _______________________________________

g. What is the effect size? () ____________________________________

h. Write out the results in APA style.

Answer 1

a. the dependent variable is the exam score

b. null hypothesis: the students who are taught the old teaching method will have the same score as the students who are taught the new teaching method

alternative hypothesis; the students who are taught the new teaching method will have a better score than the students who are taught the old teaching method

c. sample size n is 6. population mean mu is 40 sample mean is 46.

sum of squares is 28. degrees of freedom is 5 variance is 5.6 standard deviation is 2.366 standard error is 2.366/sqrt(6) = 0.966

d. t = (46-40)/0.966 = 6.21

e. critical value = 2.015 (5 DOF and alpha = 0.05)

f. if the t-value is greater than the critical value then you reject the null hypothesis. If the t value is less than the critical value then you do not reject the null hypothesis. Here, we reject the null hypothesis

g. the effect size is (46-40)/5.6 = 1.07

h. There is sufficient evidence that the new teaching method improves test scores.