A technician compares repair costs for two types of microwave ovens (type I and type II). He believes that the repair cost for type I ovens is greater than the repair cost for type II ovens. A sample of 61 type I ovens has a mean repair cost of $79.60, with a standard deviation of $23.47. A sample of 45 type II ovens has a mean repair cost of $75.25, with a standard deviation of $15.07. Conduct a hypothesis test of the technician's claim at the 0.05 level of significance. Let μ1 be the true mean repair cost for type I ovens and μ2 be the true mean repair cost for type II ovens. Step 3 of 4 : Determine the decision rule for rejecting the null hypothesis H0. Round the numerical portion of your answer to three decimal places.

The question is missing some parts. Here is the complete question.

A technician compares repair costs for two types of microwave ovens (type I and type II). He believes that the repair cost for type I ovens is greater than the repair cost in type II ovens. A samplle of 61 type I ovens has a mean repair cost of $79.60, with a standard deviation of $23.47. A sample of 45 type II ovens has a mean repair cost of $75.25, with standar deviation of $15.07. Conduct a hypothesis test of the technician's claim at the 0.05 level of significance. Let be the true mean repair cost for type I ovens and be the true mean repair cost for type II ovens.

Step 1 of 4: State null and alternative hypothesis for the test.

Step 2 of 4: Compute the value of test statistics. Round your answer to 2 decimal places.

Step 3 of 4: Determine the decision rule for rejecting the null hypothesis. Round the numerical portion of your answer to 3 decimal places.

Step 4 of 4: Make the decision for the hypothesis test. (Reject or fail to reject Null Hypothesis)

Answer and Step-by-step explanation:

First, state null and alternative hypothesis:

Use z-statistics:

where x₁ and x₂ are the means and:

Calculating value of test stattistic:

y = 3.752

z = 1.16

The test statistics is z = 1.16

The decision rule for rejecting null hypothesis is: ≤ test statistics

Using z-score table, it is possible to determine p-value, which is:

p-value = 0.877

Comparing p-value with level of significance (α = 0.05):

0.877 > 0.05

Therefore, we fail to reject the null hypothesis.

the answer is 21%.