Question

Two samples are taken with the following sample means, sizes, and standard deviations I1 = 24 D9 = 25 n = 72 m2 = 59 81 = 3 89 = 2 Estimate the difference in population means using a 94% confidence level. Use a calculator, and do NOT pool the sample variances. Round answers to the nearest hundredth. 11-12 Box 1: Enter your answer as an integer or decimal number. Examples: 3,-4,5.5172 Enter DNE for Does Not Exist, oo for Infinity Box 2: Enter your answer as an integer or decimal number. Examples: 3,-4, 5.5172 Enter DNE for Does Not Exist, oo for Infinity

Answer 1

24 3 Sample Mean 1 (X) = Sample Standard Deviation 1 (81) = Sample Size 1 (N) = Sample Mean 2 (X) = 72 25 2 Sample Standard Deviation 2 (52) = Sample Size 2 (N2) = 59 Based on the informationa provided, we assume that the population variances are unequal, so then the number of degrees of freedom is computed as follows: df (sĩ/nh - sing)? 2 = 125.889 T: +1 T +1 The critical value for a = 0.06 and df = 125.889 degrees of freedom is te = t3-a/2:1-1 = 1.898. The corresponding confidence interval is computed as shown below: Since we assume that the population variances are unequal, the standard error is computed as follows: se ni n2 32 72 22 59 + = 0.439 CI (X1 - X - te x se, X1 - X2 + te x se) (24 - 25 - 1.898 x 0.439, 24 - 25 +1.898 x 0.439) (-1.833,-0.167)

To the nearest hundreadths

$-1.83<\mu_1-\mu_2<-0.17$