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 120 pregnant mothers and finds that 3 of them smoked 21 or more cigarettes during pregnancy Test the researcher's statement at the a = 0.05 level of significance What are the null and alternative hypotheses? H: P = 0.05 versus H, P < 0.05 (Type integers or decimals. Do not round.) Because np. (1 - P.) = 5.7 < 10, the normal model may not be used to approximate the P-value. (Round to one decimal place as needed. Find the P-value P-value =(Round to three decimal places as needed.)

Answer 1

Given - n = 120 X=3 120 R = 0.025 Test Statistic z = P-P Pq Ooo 25 - 0.05 -0.028 0.05x0.95 0.0199 12o. 2 = - 1.26 P-value = P(Z< Zeal) = P(24-1-26) = P( 27 1:26) P-value = 0.1038 - By using statistical table p-value = 0.104 If you statisfied then like my Answer.