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. Test the hypothesis that Mrs. Smith and Mrs. Jones are equally effective teachers. Use a 0.10 level of significance. (Assume that student performance is approximately normal.)

The solution takes four stages:

(1) state the hypotheses,

(2) formulate an analysis plan,

(3) analyze sample data, and

(4) interpret results.

Stage 1: State the hypotheses.

The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: μ1- μ2= 0

Alternative hypothesis: μ1- μ2≠ 0

Note that these hypotheses constitute a two-tailed test.

The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Stage 2: Formulate an analysis plan. For this analysis, the significance level is 0.10.

Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Stage 3: Analyze sample data.

Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s1^2/n1) + (s2^2/n2)]

SE = sqrt[(10^2/30) + (15^2/25] = sqrt(3.33 + 9)

SE = sqrt(12.33) = 3.51

DF = (s1^2/n1+ s2^2/n2)^2/ { [ (s1^2/ n1)^2/ (n1- 1) ] + [ (s2^2/ n2)^2/ (n2- 1) ] }

DF = (10^2/30 + 15^2/25)^2/ { [ (10^2/ 30)^2/ (29) ] + [ (15^2/ 25)^2/ (24) ] }

DF = (3.33 + 9)^2/ { [ (3.33)^2/ (29) ] + [ (9)^2/ (24) ] }

DF = 152.03 / (0.382 + 3.375)

DF = 152.03/3.757

DF = 40.47

t = [ (x1- x2) - d ] / SE

t = [ (78 - 85) - 0 ] / 3.51 = -7/3.51

t = -1.99

where

s1 is the standard deviation of sample 1,

s2 is the standard deviation of sample 2,

n1 is the size of sample1,

n2 is the size of sample 2,

x1 is the mean of sample1,

x2 is the mean of sample2,

d is the hypothesized difference between the population means and

SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 40 degrees of freedom is more extreme than -1.99

that is, less than -1.99 or greater than 1.99.

We use the t Distribution Calculator to find P(t < -1.99) = 0.027, and P(t > 1.99) = 0.027. Thus, the P-value = 0.027 + 0.027 = 0.054.

Stage 4: Interpret results. Since the P-value (0.054) is less than the significance level (0.10), we cannot accept the null hypothesis.

Note:If you use this approach on an exam, you may also want to mention why this approach is appropriate. Specifically, the approach is appropriate because the sampling method was simple random sampling, the samples were independent, the sample size was much smaller than the population size, and the samples were drawn from a normal population.

Check attachment for well arranged equation.

The null hypothesis is rejected and the alternative hypothesis is upheld.

There's a significant difference between the mean scores of the students taught by Mrs Jones and the mean scores of those taught by Mrs Smith.

Step-by-step explanation:

Please kindly see the attached files for more explanation.