Within a school district, students were randomly assigned to one of two Math teachers - Mrs. Smith and Mrs. Jones. After the assignment, Mrs. Smith had 30 students, and Mrs. Jones had 25 students. At the end of the year, each class took the same standardized test. Mrs. Smith's students had an average test score of 78, with a standard deviation of 10: and Mrs. Jones' students had an average test score of 85, with a standard deviation of 15. Using the above information, answer the following question. a. State the null and the alternative hypothesis b. Calculate the degree of freedom for this estimation c. Estimate the value of the test statistic d. Estimate the confidence interval for the difference between the two student populations e. Using the Critical Value approach test the hypothesis that Mrs. Smith and Mrs. Jones are equally effective teachers. Use a 0.10 level of significance

Step-by-step explanation:

This is a test of 2 independent groups. Let μ1 be the mean score of Mrs. Smith's students and μ2 be the mean score of Mrs. Jones students.

The random variable is μ1 - μ2 = difference in the mean score of Mrs. Smith's students and the mean score of Mrs. Jones students.

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

H1 : μ1 ≠ μ2 H1 : μ1 - μ2 ≠ 0

This is a two tailed test.

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]² / (1/n1 - 1) (s1²/n1) ² + (1/n2 - 1) (s2²/n2) ²

From the information given,

μ1 = 78

μ2 = 85

s1 = 10

s2 = 15

n1 = 30

n2 = 25

df = [10²/30 + 15²/25]²/[ (1/30 - 1) (10²/30) ² + (1/25 - 1) (15²/25) ²] = 152.11/3.37883141762

df = 45

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(x1 - x2) / √ (s1²/n1 + s2²/n2)

t = (78 - 85) / √ (10²/30 + 15²/25)

t = - 1.99

d) Confidence interval = μ1 - μ2 ± z√ (s1²/n1 + s2²/n2)

Where z is the t test score for the confidence level. Since alpha = 0.1, confidence level = 1 - alpha = 1 - 0.1 = 0.9. From the t distribution table, test score at df of 45 = 1.301

z√ (s1²/n1 + s2²/n2) = 1.301√ (10²/30 + 15²/25) = 4.57

Confidence interval = (78 - 85) ± 4.57

Confidence interval = - 7 ± 4.57

e) we would find the critical value corresponding to 1 - α/2 and reject the null hypothesis if the absolute value of the test statistic is greater than the value of t 1 - α/2 from the table.

1 - α/2 = 1 - 0.1/2 = 1 - 0.05 = 0.95

The critical value is 1.679 on the right tail and - 1.679 on the left tail

Since - 1.99 < - 1.679, it is not in the rejection regions. Therefore, we would fail to reject the null hypothesis. Therefore, at 10% significance level, there is insufficient evidence to conclude that Mrs. Smith and Mrs. Jones are not equally effective teachers.