The owner of a local nightclub has recently surveyed a random sample of n = 40 c

Question

The owner of a local nightclub has recently surveyed a random sample of n = 40 customers of the club. She would now like to determine whether or not the mean age of her customers is greater than 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. Suppose she found that the sample mean was 32.2 years and the sample standard deviation was 5 years. What is the value of the test statistic used to make this determination? Round your answer to 2 decimal places.

Explanation / Answer

n = 40

= 32.2 = 5

x' = 30

The test statistic is the p value or percentage of correctness of the estimation of the sample mean.

x' = + z / n

=> 30 = 32.2 + z * 5 / 40

=> z = (30 - 32.2) 40 / 5

= -2.7828

P value from table = 0.9973 or 99.73%.

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