Question

c) W 15. Twenty-five tist wa smok five percent of women of child-bearing age smoke. A scien- anted to test the hypothesis that this is also the proportion of kers in the population of women who suffer ectopic pregnancies. modo so, the scientist chose a random sample of 120 women who had recently suffered an ectopic pregnancy. If 48 of these women turn out to be smokers, what is the p value of the test of the hypothesis Ho: p= 0.25 against H1:P +0.25 where p is the proportion of smokers in the population of women who have suffered an ectopic pregnancy?

Answer 1

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H₀ : p =0.25

H_a : p $\neq$ 0..25

$\hat p$ = x / n = 48/120=0.4

P₀ = 0.25

1 - P₀ = 1-0.25=0.75

Test statistic = z

= $\hat p$ - P₀ / [$\sqrt$P_{0 *} (1 - P₀ ) / n]

= 0.4-0.25/ [$\sqrt$(0.25*0.75) /120 ]

= 3.79

P(z >3.79 ) = 1 - P(z < 3.79) = 1-0.9999=0.0001

P-value = 2*0.0001=0.0002