Question

Independent random samples were selected from two quantitative populations, with sample sizes, means, and standard deviations given below.

n1 = n2 = 60

x1 = 125.3

x2 = 123.4

s1 = 5.7

s2 = 6.1

a) Construct a 95% confidence interval for the difference in the population means (μ1 − μ2). (Round your answers to two decimal places.) to

b) Find a point estimate for the difference in the population means.

c) Calculate the margin of error. (Round your answer to two decimal places.)

Answer 1

Sample Mean 1 (X1) = 125.3 Sample Standard Deviation 1 (81) = 5.7 Sample Size 1 (N1) = 60 Sample Mean 2 (X2) = 123.4 Sample Standard Deviation 2 (82) = 6.1 Sample Size 2 (N2) = 60 Based on the information provided, we assume that the population variances are equal, so then the number of degrees of freedom are df = ni + n2 - 2 = 60+ 60 - 2 = 118. The critical value for a = 0.05 and df = 118 degrees of freedom is te = tı-a/2;n-1 = 1.98. The corresponding confidence interval is computed as shown below: Since the population variances are assumed to be equal, we need to compute the pooled standard deviation, as follows: Sp (nı - 1)s +(n2 - 1)s; ni + n2 - 2 (60 - 1) ~ 5.72 + (60 60 + 60 - 2 1) 6.12 5.903 Since we assume that the population variances are equal, the standard error is computed as follows: 1 1 se + 8p V ni n2 5.903 1 V 60 + 60 1.078 Now, we finally compute the confidence interval: CI (#1 - X2 - te x se, X- X2+te x se) (125.3 - 123.4 - 1.98 x 1.078, 125.3 - 123.4 +1.98 x 1.078) (-0.234, 4.034) Therefore, based on the data provided, the 95% confidence interval for the difference between the population means H1 - Mz is -0.234 <H1 - M2 < 4.034, which indicates that we are 95% confident that the true difference between population means is contained by the interval (-0.234, 4.034).

2.

Point Estimate = 125.3 - 123.4 = 1.9

3.

Margin of error = tc * std error

Margin of error = 1.98 * 1.078 = 2.13

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