8 points a What is the sample proportion for the two regions Region 1 Region 2

8 points a what is the sample proportion for the two

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company one year from now is the same. ( 8 points ) a. What is the sample proportion for the two regions? Region 1 Region 2 Sample Size 95 115 Yes 72 80 Proportion 0.7579 0.6957 b. Compute the test statistic z used to test the hypothesis. z = 0.7579 0.6957 0.7238 ( 1 0.7238 ) ( 1 95 + 1 115 ) ¿ 1.0034 c. Can the firm conclude that the proportion of employees within the two regions who plan to stay with the company one year from now is not the same? Use the p-value approach and α = 0.05 to test the hypothesis stated above. P Value = 0.3153
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BUSI1013_Unit 9 4 The p-value is greater than 0.05, therefore we do NOT reject the null hypothesis. The firm could NOT conclude the hypothesis. d. Construct a 95% confidence interval for the difference of the population proportion for the two regions. ( 0.7579 0.6957 ) ± 1.96 0.7579 ( 1 0.7579 ) 95 + 0.6957 ( 1 0.6957 ) 115 0.0622 ± 0.1204 = (-0.0582, 0.183) 3. Use the data in and description of BUSI1013-Case.xls (from Unit 1 Exercise Question 6) to answer this question. (4 Points for each part; 8 points total ) a. Perform a statistical test to see whether the proportion of members who would recommend the Credit Union to their family and friends have increased as a result of the pilot. 1- Null hypothesis: There is no change in proportion of members who would recommend the Credit Union to their family and friends have increased as a result of the pilot. p1-p2=0 2- Alternative hypothesis: The proportion of members who would recommend the Credit Union to their family and friends has increased as a result of the pilot. p2-p1>0 Number of people before pilot = 290; Number of people who said yes before pilot = 185 Number of people after pilot = 290; Number of people who said yes after pilot = 262 P 1 = X 1 n 1 = 185 290 = 0.6379, P 2 = X 2 n 2 = 262 290 = 0.9034 Pooled estimate of p is X 1 + X 2 n 1 + n 2 = 185 + 262 290 + 290 = 0.7707 The test statistic z:
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BUSI1013_Unit 9 5 Z = P 2 P 1 P ( 1 P )( 1 n 1 + 1 n 2 ) = 0.9034 0.6379 0.7707 ( 1 0.7707 )( 1 290 + 1 290 ) = 7.61 The p-value corresponding to the test statistic is = 0.0000 The p-value is less than 0.0005, so therefore we reject the null hypothesis.
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