Question

A psychologist specializing in marriage counseling claims 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 250 married couples who completed her program, 188 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. (If necessary, consult a list of formulas.) o P The Bullhipothesis x S The alternative hypothesis: g 몸 D2 The type of test statistics (Choose one) DO DO 0 х The value of the test statistic (Round to at least three decimal places) ? The p-value (Round to at least three 0 decimal places) Can we reject the psychologist's dam that the proportion of married couples for whom her program can prevent divorce is at least 76%67 Yes ON

Answer 1

Solution :

Population proportion p=x/n =188/250 =0.752

Hypotheses:

Null -

$H_0: p \geq 0.76$

Alternate -

$H_1: p < 0.76$

Type of Test Statistic: Z statistic

Test statistic = (p-p0) /sq(p0(1-p0)/n)

= (0.752 - 0.76)/sq((0.76*0.24)/200)

= - 0.265

P value = 0.3955

Since p value > alpha, we failed to reject the null hypothesis

No

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