Question

A 0.01 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender? selection, the proportion of baby girls is less than 0.5. Assume that sample data consists of 66 girls in 144 ?births, so the sample statistic of StartFraction 11 Over 24 EndFraction results in a z score that is 1 standard deviation below 0. Complete parts? (a) through? (h) below. Click here to view page 1 of the Normal table. LOADING... Click here to view page 2 of the Normal table. LOADING... a. Identify the null hypothesis and the alternative hypothesis. Choose the correct answer below. A. Upper H 0?: pequals0.5 Upper H 1?: pgreater than0.5 B. Upper H 0?: pequals0.5 Upper H 1?: pnot equals0.5 C. Upper H 0?: pnot equals0.5 Upper H 1?: pless than0.5 D. Upper H 0?: pequals0.5

Answer 1

Given that, n = 144 and x = 66

=> sample proportion = 66/144 = 11/24 = 0.4583

The null and alternative hypotheses are,

H0 : p = 0.5

H1 : p < 0.5

Test statistic is,

=> Test statistic = -1.00

p-value = P(Z < -1.00) = 0.1587

We fail to reject H0 since, p-value > 0.01 significance level.