Question

A psychologist specializing in marriage counseling calms that, among all married couples, the proportion p for whom her communication program can prevent divorce is at least 76%. In a random sample of 215 married couples who completed her program, 149 of them stayed together. Based on this sample, can we reject the psychologist's claim at the 0.05 level of significance? Perform a one-tailed test. Then fill in the table below. Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. a The null hypothesis: P X $ The alternative hypothesis: The type of test statistic: Loose one) OSO DO The value of the test statistic (Round to at least three decimal places) 0 X 5 ? The p-value (Round to at least three decimal places.) Can we reject the wychologist's claim that the proportion of married couples for whom her program can prevent divorce is at least 76% Yes NO

Answer 1

Given: claim is is at least 76% p>0.76 (6 population proportion) Sample size n = 215 149 215 0.693 Sample proportion 0.05 (level of rignificance . To perform one-tailed test: (he due teiled alternative typothesis The mell hypothesis 0.76 Hoe? ( the clin) Alternative hypothesis þ> 0.76. Test statistic È p de () ~ N(0,12 for. § - 0.693 and under Ho: p=0.76 1076x 0.24 S.E (B) = (bling - so 0-0008484 0.02913 215 0.693 - 0.76 To a 2.300 0.02913

p-value = PET > To ( one - eided alternative ) - P[T> – 2:30) 0.989 p-value Since at x=o.os V p-value = 0.989 > Q0OS, the mall hypothesis . accept Ha : pc0.76 That is; We IYES reject the psychologists claun