Question

Consider the following data drawn independently from normally distributed populations: Use Table 1 X2 30.6 X1 = 25.7 022 a12 98.2 87.4 n2= 25 n1 20 a. Construct the 95% confidence interval for the difference between the population means. (Negative values should be indicated by a minus sign. Round all intermediate calculations to at least 4 decimal places and final answers to 2 decimal places.) Confidence interval is to b. Specify the competing hypotheses in order to determine whether or not the population means differ. О Но: — н2 %3D 0;B На: И1 - P2#0 Но: М — Р22 0;B На: M1 — M2 <0 Но: М - Р2S 0; На: Ру — Р2 > 0 c. Using the confidence interval from part a, can you reject the null hypothesis? Yes, since the confidence interval includes the hypothesized value of 0. No, since the confidence interval includes the hypothesized value of 0 Yes, since the confidence interval does not include the hypothesized value of 0 No, since the confidence interval does not include the hypothesized value of 0

Answer 1

a. In order to construct the confidence interval, we need to find the standard deviation of the joint sample. This can be done as follows:

-5, - , -2893

Since it has been explicitly state that these values have been drawn from a normal distribution, we can use the z table.

The critical z values for a 95% confidence interval are (-1.96,1.96).

The 95% confidence interval around the difference of means is constructed as follows:

Pr-1.96<z<1.96) = 0.95 X1 - X2 - (11-12) → Pr(-1.961 1.96) = 0.95

-4.9 - ( Pr(-1.96 <= -2) 1.96) = 0.95 2.8993

→ Prlu - H2) € (-4.9+11.96 2.8993) = 0.95 → Prat - M2) E (-39.57.29.77) = 0.95

b. The hypothesis for testing that the population means differ is:

Ho: 41 - 42 = 0, HA : 441-420

c. Using the confidence interval constructed in part a, we accept the null hypothesis as the 95% confidence interval around the difference of population means contains zero. We can conclude that the population means do not differ.