“Analyzing the Impact of E-Books on Student Test Scores: A Two-Tail Hypothesis Test”

QUESTION

Ackerman and Goldsmith (2011) found that students who studied text from printed hardcopy had better test scores than students who studied text presented on the screen. In a related study, a professor noticed that several students in a large class had purchased the e-book version of the course textbook. For the final exam, the overall average for the entire class was u= 81.7, but the sample of n = 9 students who used e-books had a mean of M = 77.2 with ((SS =392) Is the sample sufficient to conclude that scores for students using e-books were sufficiently different from scores for the regular class? Use a two-tail test with a= .01.(not .05). Please answer the question using all of the steps presented on your practice problem assignment. (null in word, alternative in words, null in symbols, alternative in symbols, critical region t, df, all steps in the analysis computing your computed t, make a decision, and give a conclusion. Do not skip around in the steps. Provide steps in an orderly steps by step manner.

ANSWER

“Analyzing the Impact of E-Books on Student Test Scores: A Two-Tail Hypothesis Test”

Introduction

In this analysis, we aim to determine whether students using e-books for studying the course textbook perform significantly differently on their final exams compared to students using printed hardcopies. We will conduct a two-tail hypothesis test with a significance level of α = 0.01 to ascertain whether the sample of nine e-book users exhibits test scores that are sufficiently different from the regular class. To conduct this test, we will follow a step-by-step approach.

Step 1: Null Hypothesis in Words (H0)

The null hypothesis states that there is no significant difference in test scores between students who used e-books and those who used printed hardcopies of the course textbook.

Step 2: Alternative Hypothesis in Words (H1)

The alternative hypothesis posits that there is a significant difference in test scores between students who used e-books and those who used printed hardcopies of the course textbook.

Step 3: Null Hypothesis in Symbols

H0: μe-book = μprinted

Where: μe-book represents the population mean test score for students using e-books, μprinted represents the population mean test score for students using printed hardcopies.

Step 4: Alternative Hypothesis in Symbols

H1: μe-book ≠ μprinted

Step 5: Critical Region (t)

Since we are conducting a two-tail test, we will split the significance level (α = 0.01) equally between the two tails, resulting in α/2 = 0.005 in each tail. We will consult the t-distribution table or a statistical calculator to find the critical t-values for α/2.

Step 6: Degrees of Freedom (df)

To calculate the degrees of freedom, we use the formula: df = n – 1 df = 9 – 1 df = 8

Step 7: Compute the Computed t-value

We will compute the t-value using the formula: t=M1−M2S12n1+S22n2t=n1S12​+n2S22​

​M1−M2​

Where: M1 = Mean test score of e-book users (77.2) M2 = Mean test score of printed hardcopy users (81.7) S1 = Variance of test scores for e-book users (392) n1 = Sample size of e-book users (9) S2 = Variance of test scores for printed hardcopy users n2 = Sample size of printed hardcopy users

Step 8: Make a Decision

After calculating the computed t-value, we will compare it with the critical t-values from the t-distribution table. If the computed t-value falls within the critical region, we will reject the null hypothesis. Otherwise, we will fail to reject the null hypothesis.

Step 9: Conclusion

Upon completing the analysis, we will provide a conclusion based on whether we reject or fail to reject the null hypothesis. If we reject the null hypothesis, it would suggest that there is a significant difference in test scores between e-book and printed hardcopy users. Conversely, if we fail to reject the null hypothesis, it would imply that there is no significant difference in test scores between the two groups.