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Employers sometimes seem to prefer executives who appear physically fit...Here are the data...c) You should also hesitate to conclude that increasing fitness causes an increase in ego strength. Explain why?

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The expert examines the increasing fitness causes as it increase ego strengths.

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A. Employers sometimes seem to prefer executives who appear physically fit, despite the legal troubles that may result. Employers may also favor certain personality characteristics. Fitness and personality are related. In one study, middle-aged college faculty who had volunteered for a fitness program was divided into low-fitness groups based on a physical examination. The subjects then took the Cattell Sixteen Personality Factor Questionnaire. Here are the data for the "ego strength" personality factor:

Low Fitness 4.99 4.24 4.74 4.93 4.16 5.53 4.12 5.10 4.47 5.30 3.12 3.77 5.09 5.40
High Fitness 6.68 6.42 7.32 6.38 6.16 5.93 7.08 6.37 6.53 6.68 5.71 6.20 6.04 6.51

(a) Is the difference in mean ego strength significant at the 5% level? At the 1% level? Be sure to state H0 and Ha.
Denote the mean ego strength for Low Fitness by and denote the mean ego strength for High Fitness by . Then we can set up null hypothesis H0 and Ha as follows.
H0: = ;
Ha:
Using Excel, we got the following table

t-Test: Two-Sample Assuming Equal Variances

Variable 1 Variable 2
Mean 4.64 6.429286
Variance 0.476385 0.185269
Observations 14 14
Pooled Variance 0.330827
Hypothesized Mean Difference 0
df 26
t Stat -8.23054
P(T<=t) one-tail 5.17E-09
t Critical one-tail 2.478628
P(T<=t) two-tail 1.03E-08
t Critical two-tail 2.778725

By the above table, we know that p-value=1.03E-08= which is less than 5% and 1%. So, at the 5% level or at the 1% level, we should reject the null hypothesis H0: = . We conclude that the difference in mean ego strength is ...

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  • "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
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