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BUSI1013_Unit 91Exercise # 9Marissa LeblancJune 10, 2018BUSI 1013 / Statistics for BusinessAlbert Wong
BUSI1013_Unit 921. Consider this hypothesis test:1.a.What is the pooled estimate of p?n1^p1+n2^p2n1+n2¿(200) (0.65)+(250) (0.70)(200+250)¿130+175450¿0.6´7b.Compute the test statistic z.z=´p−p0√p0(1−p0)n¿0.65−0.70√0.67(1−0.67)(1200+1250)¿−1.127787671¿−1.13c.What is the rejection rule using the critical value approach and α = 0.05?−α=0.05,using ztable=−1.64Reject H0if z≤−1.64d.Based on the rejection rule from c., what is your conclusion on the hypothesis?I will not reject the null hypothesis because Z is greater than -Zα
BUSI1013_Unit 93e.What is the p-value?p-value (-1.13) = 0.1292f.Use the above data to construct a 95% confidence interval for p1- p2^p1−^p2±zα/2√^p1(1−^p1)n1+^p2(1−^p2)n2¿(0.65−0.7)±1.96√0.65(1−0.65)200+0.7(1−0.7)250¿−0.13716,0.0371592.Partial responses from an employee satisfaction survey for two regions of a mid-size IT firm wererecorded in the attached BUSI1013-Two Sample Proportions.xlsxfile. These responses are a simple random sample of employee answers from the two regions to the question: Are you planning to stay with the company one year from now (Yes or No)? The firm wants to use this data to test the research (alternative) hypothesis that the proportion of employees within the two regions who plan to stay with the company one year from now is not the same. The null hypothesis is that the proportion of employees within the two regions who plan to stay with the company one year from now is the same. (
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Winter '15
Business, Null hypothesis, Statistical hypothesis testing, support staff