Question

Previously 5% of mothers smoked more than 21 cigarettes during their pregnancy. An obstetrician believes that the percentage of mothers who smoke 21 cigarettes or more is less than 5% today. She randomly selects 130 pregnant mothers and finds that 2 of them smoked 21 or more during pregnancy. Test the researcher's statement at the a=0.1 level of significance. The null and alternative hypothesis is H0=0.05 and H1<0.05.
What is the P-value?

Answer 1

Answer:

p value = 0.03514

Step-by-step explanation:

Hypotheses would be

$H_0: p = 0.05\\H_a: p$

(left tailed test at 10% level of significance)

Here p stands for the sample proportion of mothers smoked more than 21 cigarettes during their pregnancy.

Sample size =130

persons who smoked = 2

Sample proportion = $\frac{2}{130} \\=0.0154$

Assuming H0 to be true

Std error= $\sqrt{0.05*0.95/130} \\=0.01911$

p difference = -0.0346

Test statistic z=-1.81

p value = 0.03514

Since p is less than 0.10, significance level, we reject H0