Question

A union leader of a major company claims that the new incentive scheme of the contract has resulted in average weekly earnings well under $500 for all customer service workers. She takes a random sample of 16 workers and finds that their weekly earnings have an average of $475.25 and a standard deviation of $48.6. Assume the distribution of weekly earnings is normal. What is the approximate p-value (closest) for testing the appropriate hypotheses?

Answer 1

This is the left tailed test .

Test statistic = t

= ( - ) / s / n

= (475.25 - 500) / 48.6 / 16

= -2.037

Test statistic = -2.037

P-value = 0.0298