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 for type II ovens. A sample of 56 type I ovens has a mean repair cost of $⁢76.66, with a standard deviation of $⁢18.63. A sample of 75 type II ovens has a mean repair cost of $⁢72.66, with a standard deviation of $⁢22.09. Conduct a hypothesis test of the technician's claim at the 0.1 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 1: State the null and alternative hypotheses for the test.

Step 2: Compute the value of the test statistic. Round your answer to two decimal places.

Step 3: Determine the decision rule for rejecting the null hypothesis H0. Round the numerical portion of your answer to two decimal places.

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

Answer 1

As given, let the true means for repair costs for Type 1 and Type 2 ovens be μ1 and μ2 respectively. Here, we carry out 2-sample unpaired T-test as the 2 samples are not paired.

1) The null and alternative hypotheses are:

H₀: μ1 - μ2 <= 0

H_a: μ1 - μ2 > 0

2) Value of the test statistic is given by:

T = [ () ] / sqrt[(s1^2/n1) + (s2^2/n2)], where and are the sample means, s1 and s2 are the sample std deviations, and n1 and n2 are the sample sizes.

Plugging in the given values, we get

T = 1.22

3) The critical T-score for the right-tailed T-test at significance level of 0.1 and d.o.f = 56+75-2 = 129 is given by,

T-critical = T_{0.1, 129} = 1.29

Decision rule: We reject the null hypothesis if the T-statistic is greater than the critical T-score at the given significance level.

4) Decision: Since T = 1.22 < 1.29, we fail to reject the null hypothesis at the significance level of 0.1