Question

A 0.1 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 greater than 0.5. Assume that sample data consists of 91 girls in 169 births, so the sample statistic of 7 results in a z score that is 1 standard deviation above 0. Complete parts (a) through (h) below. 13 Click here to view page 1 of the Normal table. Click here to view page 2 of the Normal table. a. Identify the null hypothesis and the alternative hypothesis. Choose the correct answer below. A. Ho: p = 0.5 H :p#0.5 B. Ho: p# 0.5 Hy:p>0.5 D. Ho: p = 0.5 H1: p<0.5 C. Ho: p = 0.5 H :p>0.5

Answer 1

This is test for the binomial proportion. There are two outcomes either being boy or girl. Since n > 30 and np₀ > 10, we can use the normal approximation.

n= 169 null proportion p₀= 0.5

We want to test if the proportion for baby girl is greater than 0.5. So this is right one tailed test.

Test

a)

Ho :P 0.5

Hp > 0.5

b.

level of significance = $\alpha$

$\alpha$ = 0.1

c.

The sampling distribution used will be

Normal distribution

d)

Since we are testing if greater than null proportion

Right tailed test

e)

n = 169 x: no of girls x= 91  

$1596043628521_blob.png$ = x /n

=7/ 13 = 0.5385

Test Stat =

р ро po(1 - po)/n

Where the null proportion = 0.5

Test Stat = 1

f)

p-value = P(Z > |Test stat|)

=P(Z > 1)

=1 - P( Z < 1)

=1 - 0.84135 ....................using the normal distribution tables

p-value = 0.15866

g)

C.V. =

Za

=Z_0.1

C.V. = 1.2816

h)

The area is test stat > C.v.

Here Test stat < C.V.

we do not reject the null hypothesis at 10%. there is insufficient evidence to conclude that the proportion os girls is greater than 0.5.