Question

Question 2 *******ONLY NEED F-H********

A.) In order to compare the average heights of children in 5th grade in two different states in US., one takes a random sample of 45 students in state A and a random sample of 65 students in state B. If the two sample means are 62 inches and 55 inches and the two sample standard deviations are 12 and 10, then what would be the standard error of the difference of two sample means? B.) What would be the degree of freedom of the corresponding t distribution? C.) What would be the t-critical value for the 95% confidence interval? D.) What would be the Margin of Error for the 95% confidence interval? E.) What would be the 95% confidence interval for the difference of two population means? F.) What would be the computed test statistic for testing equality of mean heights in the two schools? G.) How would you communicate your decision using a p value approach? H.) if you assume that the two population standard deviations are equal in problem A, what would be your estimate of the population standard deviation based on your samples?

Answer 1

F)

Assuming equal variances, the pooled variance is,

Standard error of difference in means, SE =

= 4.893672873

Test statistic, t = (62 - 55) / 4.893672873 = 1.43

G)

Degree of freedom = n1 + n2 - 2 = 10 + 12 - 2 = 20

P-value = 2 * P(t > 1.43) = 0.1682

Since, p-value is greater than 0.05 significance level, we fail to reject null hypothesis H0 and conclude that there is no strong evidence of difference in population mean heights of children in 5th grade in two different states in US.

H)

Pooled variance = 636.8

Estimate of the population standard deviation of difference in means = = 25.235