At a local high school 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 smiths 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 o.10 level of significance. What statistical test should be used?

To test the hypothesis that Mrs. Smith and Mrs. Jones are equally effective teachers, we can use the Two-Sample t-test. The Two-Sample t-test is appropriate when comparing the means of two independent samples.

In this case, the two samples are Mrs. Smith's class and Mrs. Jones's class. We want to compare the average test scores of the two classes to determine if there is a significant difference in effectiveness.

The null hypothesis (H0) assumes that there is no difference in effectiveness between Mrs. Smith and Mrs. Jones, while the alternative hypothesis (Ha) assumes there is a difference.

H0: μ1 = μ2 (There is no difference in effectiveness)
Ha: μ1 ≠ μ2 (There is a difference in effectiveness)

The significance level is given as 0.10. This means the acceptable level of Type I error is 10%.

To perform the Two-Sample t-test, we need the sample means, standard deviations, and sample sizes for both Mrs. Smith's class and Mrs. Jones's class.

Mrs. Smith's class:
Sample mean (x̄1) = 78
Standard deviation (s1) = 10
Sample size (n1) = 30

Mrs. Jones's class:
Sample mean (x̄2) = 85
Standard deviation (s2) = 15
Sample size (n2) = 25

Now, we can calculate the t-statistic using the formula:

t = (x̄1 - x̄2) / sqrt((s1^2 / n1) + (s2^2 / n2))

With the provided values:

t = (78 - 85) / sqrt((10^2 / 30) + (15^2 / 25))

Next, we determine the degrees of freedom (df) using the formula:

df = (s1^2 / n1 + s2^2 / n2)^2 / ((s1^2 / n1)^2 / (n1 - 1) + (s2^2 / n2)^2 / (n2 - 1))

With the provided values:

df = (10^2 / 30 + 15^2 / 25)^2 / ((10^2 / 30)^2 / (30 - 1) + (15^2 / 25)^2 / (25 - 1))

Finally, we compare the calculated t-value with the critical t-value based on the significance level and the degrees of freedom. If the calculated t-value is greater than the critical t-value (in either tail), we reject the null hypothesis and conclude that there is a significant difference in effectiveness between the two teachers.

Note: The critical t-value can be obtained from a t-table or calculated using statistical software.

By following these steps and performing the necessary calculations, you can test the hypothesis and determine if Mrs. Smith and Mrs. Jones are equally effective teachers.