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 different from 0.5. Assume that sample data consists of 45 girls in 81 ​births, so the sample statistic of 5/9 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 α​?

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

Normal distribution

Student​ (t) distribution

χ2

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?

g. What are the critical​ value(s)?

h. What is the area of the critical​ region?

Answer 1

n = 81, x = 45

p̄ = x/n = 45/81 = 0.5556

a. Null and alternative hypothesis:

H_o : p = 0.5

H₁ : p ≠ 0.5

b. α​ = 0.01

c. Sampling distribution of the sample​ statistic:

Normal distribution

d. It is a​ two-tailed test.

e. Test​ statistic:

z =(p̄ -p)/(√(p*(1-p)/n)) = 1

f. p-value = 2*(1-NORM.S.DIST(ABS( 1, 1) = 0.3173

g. Critical​ values :

At α = 0.05, two tailed critical value, z_c = NORM.S.INV( 0.05 /2 ) = ± 1.96

h. Critical​ region:

Reject if z < -1.96 or z > 1.96