Running head: UNIT 9 EXERCISESUnit 9 ExercisesAli GharibiStatistics for BusinessProfessor Albert WongYorkville University
Unit 9 Exercises1.a: P=(200×0.65)+(250×0.7)200+250=0.67781.b: z=0.65−0.7√0.6778(1−0.67)(1200+1250)=−1.111.c: The Z critical value at α=0.05 is -1.645Reject the null hypothesis if statistic value is less than -1.6451.d: The test statistic value is -1.11 is greater than -1.645 so doesn’t reject the null hypothesis.1.e: P-value (-1.11) = 0.1331.f:(0.65−0.7)±1.96.√0.65(1−0.65)200+0.7(1−0.7)250=(−0.1355,0.0355)2.a:122127280ˆˆ0.7579;0.6957195115XXppnn2.b: p = 0.7238z12ˆˆ0.75790.6957121.00411110.7238 10.723895115pppqnn
2.c: The p-value corresponding the test is =2*(1-NORMSDIST (1.004)) =0.3153The p-value is greater than 0.05, don’t reject null hypothesis. No. The firm couldn’t 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. 2.d: 0.7579 10.75790.6957 10.69570.75790.69571.9695115= (-0.058,0.183)3.a:Null hypothesis: There is no change in proportion of members who would recommend the CreditUnion to their family and friends have increased as a result of the pilot.
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