Question

Asked Oct 15, 2019

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A newspaper article reports that the mean cost for preventative dental care per year is $175 with a standard deviation of $35 (assume that this is the population standard deviation). You want to test whether the mean cost for preventative dental care in your company is different from what was reported in the article. You take a random sample of 250 employees and find the mean cost is $171. Does this signficantly differ from what was reported in the article? Run the appropriate test at α=0.05. Give each of the following to receive full credit: 1) the appropriate null and alternative hypotheses; 2) the appropriate test; 3) the decision rule; 4) the calculation of the test statistic; and 5) your conclusion including a comparison to alpha or the critical value. You MUST show your work to receive full credit. Partial credit is available.

Step 1

**1)**

**Null and alternative hypotheses:**

*Null hypothesis:*

*H*_{0}: *µ* = 175

That is, the mean cost for preventative dental care per year is 175.

*Alternative hypothesis:*

*H*_{1}: *µ* ≠175

That is, the mean cost for preventative dental care per year is different from 175.

Since the alternative hypothesis states *µ* ≠ 175, this test is a **Two-tailed** test.

Step 2

**2)**

Here, to check whether the mean cost for preventative dental care per year is different from 175. Since it is assumed that *X *has a normal distribution with known *σ*, the standard normal is the appropriate sampling distribution. Therefore, the appropriate test is **z-test for single mean**.

Step 3

**3)**

*Decision rule:*

If* p*-value ≤ α, then reject the null hypothesis. Otherwise, ...

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