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 43 type I ovens has a mean repair cost of $70.15, with a standard deviation of $20.62. A sample of 57 type II ovens has a mean repair cost of $65.36, with a standard deviation of $22.21. 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 1 of 4: State the null and alternative hypotheses for the test.
Step 2 of 4: Compute the value of the test statistic. Round your answer to two decimal places.

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.

Step 4 of 4: Make the decision for the hypothesis test. Reject or Fail to Reject Null Hypothesis

Answer 1

Answer:

There is no enough evidence that the repair cost for type I ovens is greater than the repair cost for type II ovens.

Step-by-step explanation:

We have to perform a hypothesis test on the difference of two means.

Type I ovens:

- Sample mean: 70.15

- Sample standard deviation: 20.62

- Sample size: 43

Type II ovens:

- Sample mean: 65.36

- Sample standard deviation: 22.21

- Sample size: 57

Step 1. The claim is that the repair cost for type I ovens (μ1) is greater than the repair cost for type II ovens (μ2). The null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a: \mu_1-\mu_2>0$

Step 2.

The difference between the sample means is:

$M_d=\bar x_1-\bar x_2=70.15-65.36=4.76$

The standard error of the difference between means is:

$s_{M_d}=\sqrt{\frac{s_1^2}{n_1}+ \frac{s_2^2}{n_2}}=\sqrt{\frac{20.62^2}{43}+ \frac{22.21^2}{57}}=\sqrt{ 9.89 +8.65}=\sqrt{18.54}\\\\s_{M_d}=4.3$

The degrees of freedom are:

$df=n_1+n_2-2=43+57-2=98$

The t-statistic is

$t=\frac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\frac{4.76-0}{4.3}=1.11$

Step 3. The critical value for a right tail test with significance level of 0.05 and 98 degrees of freedom is t=1.66.

Then, if the test statistic is higher than 1.66, the effect is significant and the null hypothesis is rejected (rejection region).

If not, the null hypothesis is not rejected (acceptance region).

Step 4. The t-statistic is below the critical value, so the null hypothesis is not rejected.

There is no enough evidence that the repair cost for type I ovens is greater than the repair cost for type II ovens.