In a nationwide standardized test of workplace performance, Canadian employees s

Question

In a nationwide standardized test of workplace performance, Canadian employees score =200 on average, with =30. In a random sample of n=40 Ottawa-area employees, the mean score on the workplace performance test was equal to a z-score of 1.80. Based on this information, make a decision about the following hypotheses:

H0: Ottawa-area employees' workplace performance is equal to the performance of Canadian employees in general.

H1: Ottawa-area employees' workplace performance is not equal to the performance of Canadian employees in general. Assume that the rejection region comprises 5% of the total area of the sampling distribution of the mean of workplace performance.

(A) Select one:

a. Reject the null

b. Retain the null

(B) What is the mean score on the test in the Ottawa sample? (you will need to convert the z-score of 1.80 based on the information provided above).

Explanation / Answer

(A)For 5% level of significance the critical region is +/-1.96. Since, the test statistic 1.80 does not fall in the critical region so fail to reject the null hypothesis. Thus retain the null.

(B)The mean score on the test in the Ottawa sample is:

X=+z

=200+1.80*30

=254

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