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-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 p-valuep-value of 0.090.09 . 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

The probability is 0.10.

Step-by-step explanation:

Hello!

When conducting a test there are four different decisions you can make.

1. Reject the null hypothesis when the hypothesis is false (This is a correct decision) (True positive, TP)

2. Reject the null hypothesis when the hypothesis is true (This decision is also known as Type I error) (False positive, FP)

3. Fail to reject the null hypothesis when the hypothesis is true (This is a correct decision) (True negative, TN)

4. Fail to reject the null hypothesis when the hypothesis is false (This decision is also known as Type II error) (False negative, FN)

Each of these decisions has an associated probability:

1. P(TP)= 1-β (Known as the power of the test)

2. P(FP)= α (Known as significance level)

3. P(TN)= 1-α

4. P(FN)= β

With this in mind, if the tested null hypothesis is "The new equipment has no effect in the jewelry making production" vs the alternative hypothesis "The new equipment increases the jewelry making production" and Frances rejects the null Hypothesis when this is in fact, true, then the decision made is a Type I error and its probability is α= 0.10

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europa2433

europa2433

step-by-step explanation:

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