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 $471.45 and a standard deviation of $41.6. Assume the distribution of weekly earnings is normal. What is the approximate p-value (closest) for testing the appropriate hypotheses?
This is the left tailed test .
Test statistic = t
= ( - ) / s / n
= (471.45 - 500) / 41.6 / 16
= -2.745
Test statistic = -2.745
df = 15
P-value = 0.0075
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