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.

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Asked Oct 14, 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.

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Expert Answer

Step 1

Given:

Sample size=250

Population standard deviation=$35

Sample mean=$171

Significance level=0.05

Explanation:

To test whether the mean cost for preventative dental care in your company is different from what was reported in the article.

Hypothesis test:

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Null Hypothesis : u =175 Alternative Hypothesis : i#175

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Step 2

Hence, the null and alternative hypothesis is:

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Null Hypothesis : u =175 Alternative Hypothesis : i#175

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Step 3

2)  The appropriate test:

See the hypothesis test condition, the condition of hypothesis test is two tailed test.

3) To explain the decision rule.

Explanation:

We reject the null hypothesis and accept the alternative hypothesis if z-calculate is greater than z-critical and accept the null hypothesis and reject the alternative ...

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х — и 171-175 35 250 =-1.8070 l=1.807

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