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Hypothesis Testing Research And Evaluation Ii Res 342

1576 words - 7 pages

Hypothesis TestingLaura is the Human Resources manager of a major company, MasterCard. MasterCard is opening up three new facilities and would like Laura to consider relocating to Arizona, California or Illinois. Laura got a team together to help figure out which location would be more efficient. The team has to do research on pay and cost of living expenses. Since transportation is a necessity in today's fast paced environment and a valuable factor to consider when relocating we decided this would be our first step is the hypothesis testing. "Hypothesis testing starts with a statement, or assumption, about a population parameter - such as the population mean. It is a procedure based on ...view middle of the document...

To determine the null you use the Appendix D chart. Since the level of significance is .05 and it is a two tail test you should divide .05 by 2. You then take the area of ½ of the chart (.5000) minus the .025 and get .4750. You look for that number on the chart and get your null that way.The second step is to select the level of significance. We chose to use the 90% confidence interval which is .05 because it is usually selected for consumer projects so it would help give us more accurate results for our decision.The third step is to select the test statistics. A test statistic is "a value, determined from sample information, used to determine whether to reject the null hypothesis" (Lind et al., 2004, p. 321). For the purpose of this paper we are using the "t" and "z" test statistic. For the "t" test we are going to take a random 25 sample to keep it under the 30 cut off point. Then to get the results of the "z" test we will take the entire sample (greater than 30) to determine the "z" test statistics.Step four is formulating the decision rule. To formulate the decision rule the team must first, find the critical value of "z", which is the reject/not reject cutoff. This is going to be a two-tailed test with a LOS of 0.05 in which the team is trying to prove that the sample mean is equal to the population mean, therefore the critical value of z is -1.65. The decision rule is: Reject and accept if the calculated value of z is equal to -1.65.The final step is making the decision. This is done by computing the test statistic and comparing the critical value to it. From this you determine to reject or accept the null. Look at the information and test statistics below to see if we rejected or accepted the null.We are going to show each step in the "t" and "z" test.Step 1: State the null and the alternate hypothesis= 1.708≠ 1.708Step 2:Select the Level of Significance.05Step 3:Identify the Test StatisticThe sample size is less than 30 so we are going to use the "t" calculation.Step 4:Formulate the decision rule.To do this we are going to use the formulat = 2.9836 - 2.99 = -.0064 = -0.34.095258 ∕√25 .01905Step 5: Take the Sample and arrive at the decision.Step 1: State the null and the alternate hypothesis= 1.65≠ 1.65Step 2:Select the Level of Significance.05Step 3:Identify the Test StatisticThe sample size is more than 30 so we are going to use the "z" calculation.Step 4:Formulate the decision rule.To do this we are going to use the formulaz = 2.91175-2.99 = -.07825 = -8.88.05570 ∕√40 .00881Step 5: Take the Sample and arrive at the decision.We are going to add another step to our portion of the test. We are going to come up with the P-value of each statistic. The "t" statistic and the "z" statistic had different methods in determining the P-value. To determine the P-value in the z test we are again going to use Appendix D and find the value of 1.65 which is .4505. We are then going to subtract .5000 - .5000 and come...

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