Question

A recent study of a random sample of employees evaluated whether satisfaction is related to job performance. 250 employees who have been employed greater than 10 years, and 250 employees who have been employed less than 10 years were included in the study. The number of employees who still worked for the company after this study were 236 and 226, respectively. Assume that the two groups are independent, and that satisfaction rates follow a normal distribution in the population. Use a 0.05 significance level to test the claim that the rate of satisfaction among those employed less than 10 years is greater than those employed more than 10 years. What is your decision about the null hypothesis?

A. Fail to reject the null. There is sufficient evidence to support the claim that those employed less than 10 years have a higher level of satisfaction than those employed greater than 10 years.

B. Reject the null. There is sufficient evidence to support the claim that those employed have a higher level of satisfaction than those employed less than 10 years.

C. Reject the null. There is sufficient evidence to support the claim that the level of satisfaction of those employed less than 10 years is the same as that for those employed more than 10 years.

D. Fail to reject the null. There is sufficient evidence to support the claim that the level of satisfaction of those employed less than 10 years is the same as for those employed more than 10 years.

E. Reject the null. There is insufficient evidence to support the claim that for those employed less than 10 years is greater than for those employed more than 10 years.

Answer 1

Ho:   p1 - p2 =   0          
Ha:   p1 - p2 >   0          
                  
sample #1   ----->   experimental          
first sample size,     n1=   250          
number of successes, sample 1 =     x1=   236          
proportion success of sample 1 , p̂1=   x1/n1=   0.9440          
                  
sample #2   ----->   standard          
second sample size,     n2 =    250          
number of successes, sample 2 =     x2 =    226          
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.904          
                  
difference in sample proportions, p̂1 - p̂2 =     0.9440   -   0.9040   =   0.0400
                  
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.9240          
                  
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.0237          
Z-statistic = (p̂1 - p̂2)/SE = (   0.040   /   0.0237   ) =   1.6876
                  

p-value =        0.0457   [excel function =NORMSDIST(-z)]      
decision :    p-value<α,Reject null hypothesis               
                  
Conclusion: Those whoc have less than 10 years have high satisfaction than those who have greater than 10 years satisfaction.

THANKS

revert back for doubt

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