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

Question

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)

Answer 1

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