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Setting Up For Standard Deviation

The end of the year can be a time of rejoicing, a time of business, and a time of confusion. For most employers, it is a time to conduct open enrollment for its employees. For employers and employees, this can be a very confusing time with all of the options made available for the health care industry. The summary detailed below will evaluate the differences between an in-network dental provider and an out of-network dental provider.

An employee could face a daunting experience of open enrollment for employee benefits; if the employee is not properly educated about the benefits that are available to him or her. As this is the time for most employers to offer open enrollment, our team felt this would be a good topic to explore. Our team is looking specifically at a dental plan offered by a team member's employer. The dental plan is through Met Life Dental. Met Life Dental offers an in-network and an out of-network dental plan. Our team is going to show how staying in-network will save an employee up to fifty-eight percent more money than using a dentist out of-network.

An in-network provider and the participating insurance companies have negotiated the amount of services to be charged to a participating member. An out of-network provider has reasonable and customary charges and a percentage of those charges is attached to the amount that is owed by the patient, resulting in an increase over what an in-network provider would charge. We will provide evidence of this through the use of the five-step procedure for testing a hypothesis.
The first step is to state the null hypothesis which is H0: u = &#8805; 58%. The second step is to state the alternate hypothesis which is H1: u &#8800; <58% and this will be a one-tailed test. The third step is to select the level of significance and due to this is a consumer research project our team has selected .05 level.

The forth step is to select the test statistic and our team is going to use a t distribution. The reason the team chose to use a t distribution because our sample size is small with four observations. The value of the test statistic is computed by t = X - µ s/&#8730;n. The next step is to formulate the decision rule. We used the table in the text book of Appendix F to assist in formulating the decision rule. The number of degrees of freedom is the total number of observations in the sample minus the number of samples. In this example, we have 4 observations in the sample minus one = three degrees of freedom. To locate the critical value, we located the row with the appropriate degrees if freedom and since this is a one-tail test, we were able to locate the level of significance is .025. To find the critical value you locate the level of significance and move down the column to where the degrees of freedom intersect and we find that the critical value is 3.182. Because this is a one-sided test and the rejection region is left tailed, the critical value will be negative. The decision rule than is to reject H0 if the value of t is less than -3.182.

Finally, the fifth step is to make a decision and interrupt the results.
t = X - µ s/&#8730;n = 58 - 100/??&#8730;4 I need your help. I cannot determine the standard deviation. Maybe I am setting up the problem wrong. Please let me know.

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