Question

Assume that 200 babies are born to 200 couples treated with the XSORT method of gender selection that is claimed to make girls more likely. Preliminary results from a test of the XSORT method of gender selection involved 200 couples who gave birth to 115 girls and 85 boys. At significance level of 0.1, test whether the proportion of girl births after the treatment is greater than the 50% that occurs without any treatment (optional: show your work of the hypothesis testing in paper). Answer the next three questions
a. test statistic =
b. p-value =
c. What is your decision?

Answer 1

Answer:

a) test statistic = 2.12

b) p-value = 0.017

c) we reject the Null hypothesis

Step-by-step explanation:

Given data :

N = 200

girls (x) = 115 , Boys = 85

p = x / n = 115 / 200 = 0.575

significance level ( ∝ ) = 0.1

aim : test whether the proportion of girls births after the treatment is greater than 50% that occurs without any treatment .

A) Determine the test statistic

H0 : p = 0.5

Ha : p > 0.5

to determine the test statistic we will apply the z distribution at ( ∝ ) = 0.1

Z - test statistic = ( 0.575 - 0.5)  / $\sqrt{0.5*0.5 / 200}$  = 2.12

b) determine the p-value

The P-value can be determined using the normal standard table

P-value = 1 - p(Z< 2.12 )  =  1 - 0.9830 = 0.017

c) Given that the p value ( 0.017 ) < significance level ( 0.1 )

we will reject the H0 because there is evidence showing that proportion of girls birth is > 50%