Question

A 0.05 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 78 girls in ​169 births, so the sample statistic of 6/13 results in a z score that is 1 standard deviation below 0. Complete parts​ (a) through​ (h) below.

a. Identify the null hypothesis and the alternative hypothesis:

b. What is the value of a?

a=___

c. What is the sampling distribution of the sample statistic?

__Normal distribution

__x^2

__Student (t) distribution

d. is the test two-tailed, left-tailed, or right-tailed?

e. what is the value of the test statistic?

f. What is the P-value? (Round four decimals)

g. What are the critical value/s? (Round two decimals)

h. What is the area of the critical region? (Round two decimals)

Answer 1

Given : n=169 , x=78

The estimate of the sample proportion is , $\hat{p}=\frac{x}{n}=\frac{78}{169}=0.4615$

a. The null and alternative hypothesis is ,

$H_0:p=0.5;H_1:p<0.5$

b. The value of a is a=0.05

c. The sampling distribution of the sample statistic is Normal distribution

d. The test is left-tailed test

e. The value of the test statistic is ,

Zstat P- Po po(1-po) -1 0.4615 – 0.5 0.5(1-0.5) 169 n

f. The p-value is ,

p-value=$P(Z>|Z_{stat}|)=P(Z>1)=1-P(Z\leq 1)=1-\Phi(1)$

$=1-0.8413=0.1587$ ; From standard normal distribution table

g. The critical value is ,

; From Z-table

Za 20.05 -1.64

h. The area of the critical region is $(-\infty ,-1.64]$