Question

4. A woman who smokers during pregnancy increase health risks to the infant so that it is known that about 20% (1/5) of female smokers quit smoking during pregnancy. Suppose that a random sample of 400 pregnant women who smoked prior to pregnancy contained 62 who quit smoking during pregnancy.
a. What is the point estimator for p, the population of female smokers who quit smoking during pregnancy?

b. What are the assumptions (requirements) for constructing a confidence interval for p?

c. Compute the 95 % confidence interval for p.

d. Find the margin of error from part c.

e. Interpret the confidence interval you found in part c.

f. Does the result from c) agree with the claim saying, “about 20% (1/5) of female smokers quit smoking during pregnancy”?

Answer 1

Solution :

a) Point estimate = sample proportion = = x / n = 62/400 = 0.155

b) n = 400

x = 62

Point estimate = sample proportion = = x / n = 62/400 = 0.155

1 - = 1-0.155 = 0.845

b) At 95% confidence level

= 1-0.95% =1-0.95 =0.05

/2 =0.05/ 2= 0.025

Z/2 = Z0.025 = 1.960

Z/2 = 1.960

c) Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.960 * ((0.155*(0.845) / 400)

= 0.035

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.155 -0.035 < p < 0.155 +0.035

0.120< p < 0.190

( 0.120 , 0.190 )

d) A 95% confidence interval for population proportion p is =( 0.120 , 0.190 )

e)

This is the right tailed test .

The null and alternative hypothesis is

H0 : p = 0.20

Ha : p > 0.20

= x / n = 62/400 = 0.155

P0 = 0.20

1 - P0 = 0.80

Test statistic = z

= - P0 / [P0 * (1 - P0 ) / n]

=0.155 -0.20 / [0.20*(0.80) /400 ]

= -2.487

P(z >-2.487 ) = 1 - P(z < -2.487 ) = 0.9936

P-value = 0.9936

= 0.05

0.9936 > 0.05

Do not reject the null hypothesis .

There is insufficient evidence to suggest that about 20% (1/5) of female smokers quit smoking during pregnancy