Question

Question 3: Behavioural researchers have developed an index designed to measure man- agerial success. This index is assumed to be following a normal distribution. A researcher wants to compare the average success index for two groups of managers at a large manu- facturing plant. First group of managers (F) engage in a high volume of interactions with people outside the manager's work unit. Second group of managers (S) rarely interact with people outside their work unit. It is assumed that the two groups have the same varianoe in success index. The data are: 65 58 78 60 68 69 66 70 53 71 62 63 66 50 57 68 53 60 68 54 The following is the MINITAB output for the summary statistic: Variable N Mean Median TrMean StDev SE Mean 10 65.80 57.00 65.87 7.21 2.28 10 60.1061.00 60.38 6.42 2.03 Variable Minimum Maxiu 3 53.00 78.00 59.50 70.25 50.00 68.00 53.75 66.50

1. Obtain a 95% confidence interval for the difference in mean success indexes of managers in the two group.

2. Does the interval in part (a) support the statement “on average, the success index for the two groups of managers are significantly different”? Why?

3. At 5% level of significance, do that data support the argument that the first group has a significantly higher index than the secon group?

Can someone explain this with detail and formulas?

Answer 1

$Since~population~variances~are~equal~but~unknown~so~we~estimate~the`population~variances~by\\ s^2=\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}=46.6003\\ where,~n_1=n_2=10,~s_1=7.21,~s_2=6.42.\\ t_{0.025,18} =2.1009,~\bar{x}_1=sample~mean~of~F=65.80,\\~\bar{x}_2=sample~mean~of~S=60.10\\ Then~95\%~C.I.~for~the~difference:\\ \left(\bar{x}_1-\bar{x}_2-t_{0.025,18}s\sqrt{\frac{1}{n_1}+\frac{1}{n_2}} ,~\bar{x}_1-\bar{x}_2+t_{0.025,18}s\sqrt{\frac{1}{n_1}+\frac{1}{n_2}} \right )=(-0.7138,~12.1138).\\ Since~the~interval~contains~zero~hence~the~the~interval~does~not~support~the~statement.$

$Null~hypothesis,~H_0:\mu_d=0~vs.~Alternative~hypothesis,~H_1:\mu_d>0\\ Test~statistic=t=\frac{\bar{x}_1-\bar{x}_2}{s\sqrt{\frac{1}{n_1}&plus;\frac{1}{n_2}}}=1.8671\\ Critical~value=t_{0.025,18}=2.1009\\ Since~value~of~t<Critical~value~so~we~fail~to~reject~null~hypothesis~and~conclude~that~\\$

data does support the argument that the first group has a significantly higher index than the second group.