Question

Suppose Frances is a researcher at Beaded Gemsz, a company that makes beaded jewelry. She wants to evaluate whether using better equipment during the current year has increased jewelry-making productivity. The company's reporting team estimated an average daily production yield of 105 units per store from previous years.

Frances conducts a one-sample ?-z-test with a significance level of 0.100.10, acquiring daily unit yield data from each of the stores' databases for 45 randomly selected days of the year. She obtains a ?-value of 0.09 The power of the test to detect a production increase of 12 units or more is 0.980.98.

What is the probability that Frances concludes that the new equipment increases the average daily jewelry production when in fact the new equipment has no effect? Provide your answer in decimal form, precise to two decimal places.

Answer 1

Here the null hypothesis is that the productivity is unchanged, and the alternative is that the productivity is increased. The level of significance is a =0.10. The P-value is 0.09. The power of the test is 0.98. The type I error is defined as rejecting the null hypothesis when actually it is true. Here rejecting the null hypothesis means that the productivity is increased. Now, the probability that Frances concludes that the new equipment increases the average daily jewelry production when in fact the new equipment has no effect is calculated as follows: P(productivity increased | Not effect)=P(Reject HH.) =P(Type I error) = a = 0.10