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 78 girls in 169 ​births, so the sample statistic of six thirteenths 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. Choose the correct answer below. a) H0: p does not equal 0.5 / H1: P>0.5 b.) H0: p =0.5 / H1: p does not equal 0.5 c) H0: p = 0.5 / H1: p> 0.5 d) H0: p=0.5 / H1: p<0.5

B. What is the value of a?

C. What is the sampling distribution of the sample statistic? a. X^2 b. Student (t) distribution c. Normal 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?

G. What are the critical value(s)?

H. What is the area of the critical region?

Answer 1

Answer)

A)

Null hypothesis Ho : P = 0.5

Alternate hypothesis Ha : P does not equal to 0.5

B)

Alpha = 0.01

C)

N = 169

P = 0.5

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 84.5

N*(1-p) = 84.5

Both the conditions are met so we can use standard normal z table to estimate the P-Value

So answer is normal distribution

D)

Two tailed

E)

Test statistics z = (oberved p - claimed p)/standard error

Standard error = √{claimed p*(1-claimed p)/√n

Claimed p = 0.5

N = 169

Observed P = 78/169

Test statistics z = -1

F)

From z table, P(z<-1) = -0.1587

As our test is two tailed so, P-value = 2*0.1587 = 0.3174

G)

alpha = 0.01

As the test is two tailed

So first we will divide 0.01 into two parts

0.005

From z table, P(z<-2.58) = P(z>2.58) = 0.01

So critical values are -2.58 and 2.58

H)

Rejection region is

Reject Ho if

Test statistics is > 2.58

Or < -2.58