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 0.08 , acquiring daily unit yield data from each of the stores' databases for 45 randomly selected days of the year. She obtains a P -value of 0.06 . The power of the test to detect a production increase of 12 units or more is 0.85 . 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

There is 8% (P=0.08) that Frances concludes that the new equipment increases the average daily jewelry production when in fact the new equipment has no effect.

Step-by-step explanation:

We have one-sample z-test with a significance level of 0.08 and a power ot the test of 0.85.

In this test, the null hypothesis will state that the new equipment has the same  productivity of the older equipment. The alternative hypothesis is that there is a significative improvement from the use of new equipment.

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 equal to the probability of making a Type I error (rejecting a true null hypothesis).

The probability of making a Type I error is defined by the level of significance, and in this test this value is α=0.08.

Then, there is 8% that Frances concludes that the new equipment increases the average daily jewelry production when in fact the new equipment has no effect.

the graph top, left is y = 4x

its the fourth option