28) According to CNN business partner Careerbuilder.com, the average starting sa

Question

28) According to CNN business partner Careerbuilder.com, the average starting salary graduates in 2008 was at least $47.413, Suppose that the American Accountants planned to test this claim by randomly sampling 200 accoun who graduated in 2008. Compute the power of the hypothesis test to reject the null hypothesis if the true avera salary is only $47,000. Assume that the population standard deviation is know to the test is to be conducted using an alpha level equal to 0.01. be $4,600 and A) 0.1323 B) 00872 C) 0.8554 D) 09812 29)_ to an industry report, 26 percent of all households have at least one cell phone. Further, 90 with a standard deviation equal 9.60. Recently, a random sample of 400 households was selected. Of these households, 88 indicated that they had cell phones. The mean bill for these 88 households was $57.00. What is the probability of getting 88 or fewer households industry report are correct? with cell phones if the numbers provided by the A) Nearly 0.4656 B) About 0.1345 C) Approximately 0.0344 D) Can't be determined without knowing the standard deviation 30) 30) The St Joe Company grows pine trees and the average annual increase in tree diameter is 31 inches with a standard deviation of 0.5 inch. A random sample of m 50 trees is collected. What is the probability of the sample mean being less the 2.9 inches? A) 0.4977 B) 0.9977 C) 00023 D) 09954

Explanation / Answer

Solution Q28

Let X = Starting salary of accounting graduates with mean µ and standard deviation ?. Then, in the given situation,

H0: µ = 47413 Vs H1: µ < 47413.

The test statistic (with ? known) is:

Z = ?n(Xbar – 47413)/?

= (?200)(Xbar – 47413)/4600.

Given ? = 0.01, null hypothesis is rejected if Z < Z0.99 = - 2.32635

i.e., Xbar > 46656.3102

Now, given actual µ = 47700,

P(null hypothesis is rejected) = P(Xbar > 46656.3102/µ = 47700, ? = 4600 and n = 200)

= P(Z > - 1.0566)

= 0.8547 ANSWER

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