Question Help In a recent year, the average hourly compensation far federal gove

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Question Help In a recent year, the average hourly compensation far federal govemment employees was $40.32. The accompanying data are a random sample of the hourly amounts paid to local government emplayees. Complete parts a through d below Click the icon to view the hourly wages. a. If hourly pay is thought to be nomally distributed, what proportion of local government employees receive pay higher than the average for federal employees? Round to two decimal places as needed.) b. Calculate the sample proportion of the lccal emplayees whose haurly compensation is less than $40.32 Round to two decimal places as needed.) c. Detemine the probability that the sample proportion would be equal to or greater than the answer cbtained in part b The probability is Round to four decimal places as needed.) d. On the basis of the work in parts a, b, and , is it reasnable to conclude that the proportion of state employees whose hourly compensation is less than $40.32 is the same as that of the federal employees? Explain The prabability that the sample proportion would be equal to or greater than the answer abtained in part b is | 0.05, so it is I to conclude that the proportion of state employees whose hourly compensation is less than $40.32 is the same as that of the federal employees. Hourly Wages 41.28 3028 42.39 40.82 39.19 34.39 36.44 3887 33.68 40.79 39.51 49.87 23.64 35.02 44.22 30.38 40.42 47.97 33.31 31.32 25.83 28.13 35.36 36.01 42.23 36.17 27.38 34.68 46.79 38.77 52.76 4148 29.51 38.04 4291 3B.62 32.82 4588 29.47 45.92 38.76 34.29 23.09 35.17 31.52 44.26 3081 30.84 34.97 35.39 25,72 34.25 3483 38.47 39.66 2939 35.04 28.87 37.17 30.22 PrintDone nter your answer in each of the answer boxes.

Explanation / Answer

a. Here if hourly pay is normally distributed then proportion of local government employees receive pay higher than the average of fedral employees = 0.50

b .There are 44 employees who got wages less than $ 40.32.

Proportion of employees higher wages than average of fedral employees = 44/60 = 0.7333

c. Standard error of the proportion se = sqrt(0.5 * 0.5 /60) = 0.06455

Here Pr(p > 0.73 ; 0.50 ; 0.06455) = 1 - NORM(p <0.73; 0.50; 0.06455)

Z = (0.73 - 0.50)/ 0.06455 = 3.56

Pr(Z < 3.56) = 0.9998

Pr(p > 0.73 ; 0.50 ; 0.06455) = 1 - NORM(p <0.73; 0.50; 0.06455) = 1 - 0.9998 = 0.0002

d. so based on the results

p = 0.0002 and we reject the null hypothesis and can conclude that local employees receive significantly lower compensation than the average of federal employees.

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