Question

In: Statistics and Probability

A professor decides to conduct a experiment to determine the effects of using a study sheet...

A professor decides to conduct a experiment to determine the effects of using a study sheet on final performance among her students. She is interested in finding out if creating a study sheet improves performance on the final. Prior to the start of the semester, the professor randomly assigns 16 students to one of two groups. One group of students is given a direction to create a study sheet over the course of the term and the other group is not. Both groups receive the same course content over the course of the semester. Scores from the final are used as the dependent variable. Below are the scores for the final of students in her class.

Study Sheet: 88, 77, 96. 85, 71, 73, 81, 91
No study sheet: 63, 71, 83, 90, 92, 84, 72, 71

a.  Is this an independent samples or related samples design? Why?
b.  Write the H0 and H1 in symbols.
c,  Calculate the degrees of freedom (df) and the t critical value with a significance level of .05.
d.  Use the data and conduct the appropriate test to test the hypothesis that creating a study sheet will improve performance on the final
e. Report your decision.
f. Interpret your finding.

Solutions

Expert Solution

a) This is an independent samples design as students are divided into two different groups.

b) Hypothesis

c)

Degrees of freedom = n1 + n2 -2 = 8+8-2 = 14

t critical value at significance level of 0.05 and df=14 is 2.144 (by referring to the two-tailed t table

D) Following formula can be used to calculate the t statistic:

where,

x1 = group 1 mean

x2 = group 2 mean

s1 = group 1 standard deviation

s2 = group 2 standard deviation

We need to calculate the mean and standard deviation for each group:

Excel can be used to calculate the mean and standard deviation.

Mean can be calculated by using = average command

Standard deviation can be calculated by using =stdev.s command

x1 = 82.75, x2 = 78.25 , s1 = 8.82 s2 = 10.41

By putting the given values in the above formula, we get:

t = 0.9275

the p-value at t = 0.9275 and df = 14 is 0.3694

Since the calculated p value 0.3694 is more than the significance level of 0.05. We accept the null hypothesis.

There is no difference in performance between the students having study sheet or no study sheet


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