Question

An experiment was conducted to compare the use of iPads versus regular textbooks in teaching algebra to two classes of middle school students.† To remove teacher-to-teacher variation, the same teacher taught both classes, and all teaching materials were provided by the same author and publisher. Suppose that after 1 month, 10 students were selected from each class and their scores on an algebra advancement test recorded. The summarized data follow.
iPad Textbook
Mean 86.8 79.5
Standard Deviation 8.97 10.8
Sample Size 10 10

Required:
a. use the summary data to test for a significant difference in advancement scores for the two groups using α= 0.05.
b. Find a 95% confidence interval for the difference in mean scores for the two groups.
c. In light of parts (a) and (b), what can we say about using iPads versus traditional textbooks in teaching algebra at the middle school level?

Answer 1

Answer:

Step-by-step explanation:

Hello!

The objective is to determine if there is any difference between using iPads vs textbooks in teaching algebra.

Two middle school classes were selected, to eliminate any other source of variation, the same teacher taught both classes, and the materials were provided by the same author and publisher. After a month 10 students of each class were randomly selected and tested, their test scores were recorded:

X₁: test scores of students that used iPads to study.

n₁= 10

X[bar]₁= 86.8

S₁= 8.97

X₂: test scores of students that used regular textbooks to study.

n₂= 10

X[bar]₂= 79.5

S₂= 10.8

a.

H₀: μ₁=μ₂

H₁: μ₁≠μ₂

α:0.05

Assuming that both variables are normally distributed and the population variances are equal, the statistic to use is a Student t for two independent samples with pooled sample variance:

$t_{H_0}= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$

$Sa^2= \frac{(n_1-1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2}$

$Sa^2= \frac{9*80.4609+9*116.64}{10+10-2} = 98.55$

Sa= 9.93

$t_{H_0}= \frac{86.8-79.5}{9.93\sqrt{\frac{1}{10} +\frac{1}{10} } } = 1.64$

p-value: 0.118364

The p-value is greater than the significance level so the decision is to not reject the null hypothesis. This means that there is no significant evidence between the scores of the two groups.

b.

95% CI

(X[bar]-X[bar])±$t_{n_1+n_2-2;1-\alpha /2}$*$Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }$

$t_{n_1+n_2-2;1-\alpha /2}= t_{18;1.975}= 2.101$

(86.8-79.5)±2.101*(9.93$\sqrt{\frac{1}{10} +\frac{1}{10} }$)

[-2.03; 16.63]

With a 95% confidence level, you'd expect that the interval [-2.03; 16.63] would contain the difference between the mean scores of the two classes.

c.

Considering that the null hypothesis wasn't rejected and that at the same level the confidence interval includes the zero, we can affirm that the format of the teaching materials, digital or regular textbooks, has no significant effect on the scores of the students.

I hope it helps!