Answer to Question #88849 in Statistics and Probability for Ranjini Naashna Goundar

Question #88849
A telephone company representative estimates that 20% of its customers have call-waiting service. To test this hypothesis, she selected a sample of 100 customers and found that 37% had call waiting. At 5% significance level, is there enough evidence to reject the claim? To give your conclusion,

a. Use the traditional method. (4 marks)

b. Use the p-value method. (3 marks)

c. Use the confidence interval method. (3 marks)
1
Expert's answer
2019-04-30T10:59:13-0400

a.    Test  statistic:  z=0.370.200.2(10.2)100=4.25.Critical  zvalue:  zc=1.96.Since  test  statistic  is  greater  than  the  critical  value,  we  should  conclude  that  there  is  enough  evidence  to  reject  the  claim.b.    Pvalue  for  z=4.25:  p<0.0001.Since  Pvalue  is  less  than  0.05,  we  should  conclude  that  there  is  enough  evidence  to  reject  the  claim.c.    95CI=(0.371.960.2(10.2)100,  0.37+1.960.2(10.2)100)=(0.29,  0.45).Since  0.2  does  not  lie  than  in  the  confidence  interval,  we  should  conclude  that  there  is  enough  evidence  to  reject  the  claim.a.\;\;Test\;statistic:\;z=\frac{0.37-0.20}{\sqrt{\frac{0.2*(1-0.2)}{100}}}=4.25.\\ Critical\;z-value:\;z_c=1.96.\\ Since\;test\;statistic\;is\;greater\;than\;the\;critical\;value,\;we\;should\;conclude\;that\;\\there\;is\;enough\;evidence\;to\;reject\;the\;claim.\\ b.\;\;P-value\;for\;z=4.25:\;p<0.0001.\\ Since\;P-value\;is\;less\;than\;0.05,\;we\;should\;conclude\;that\;\\there\;is\;enough\;evidence\;to\;reject\;the\;claim.\\ c.\;\;95CI=(0.37-1.96\sqrt{\frac{0.2(1-0.2)}{100}},\;0.37+1.96\sqrt{\frac{0.2(1-0.2)}{100}})=(0.29,\;0.45).\\ Since\;0.2\;does\;not\;lie\;than\;in\;the\;confidence\;interval,\;we\;should\;conclude\;that\;\\there\;is\;enough\;evidence\;to\;reject\;the\;claim.\\


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