Previously, 5% o mothers smoked more than 21 cigarettes during their pregnancy.

Question

Previously, 5% o mothers smoked more than 21 cigarettes during their pregnancy. An obstetrician believes that the percentage o mothers who smoke 21 cigarettes or more is less than 5% today. She randomly selects 155 pregnant mothers and finds that 2 of them smoked 21 or more cigarettes during pregnancy. Test the researchers statement at the ? = 0.1 level of significance. What are the null and alternative hypotheses? Type integers or decimals. Do not round.) Find the P-value P-value(Round to three decimal places as needed.) Is there sufficient evidence to support the obstetrician's statement? O A No reject the null hypothesis because the P-value is ess than ?. T ere is not suffic ent evidence to conclude that the percentage of others who smoke 21 or more c garettes during pregnancy is less O B. Yes, do not reject the null hypothesis because the P-value is greater than a. There is sufficient evidence to conclude that the percentage of mothers who smoke 21 or more cigarettes during pregnancy O C. No, do not reject the null hypothesis because the P-value is greater than a. There is not sufficient evidence to conclude that the percentage of mothers who smoke 21 or more cigarettes during O D. Yes, reject the null hypothesis because the P-value is less than a. There is sufficient evidence to conclude that the percentage of mothers who smoke 21 or more cigarettes during pregnancy is less than 5% is less than 5%. pregnancy is less than 5%. than 5%. lick to select your answer(s) and then click Check Answer.

Explanation / Answer

The statistical software output for this problem is:

One sample proportion summary hypothesis test:
p : Proportion of successes
H0 : p = 0.05
HA : p < 0.05

Hypothesis test results:

Hence,

Ho: p = 0.05 versus H1: p < 0.05

P - value = 0.017

Option D is correct.

Proportion Count Total Sample Prop. Std. Err. Z-Stat P-value p 2 155 0.012903226 0.017505759 -2.1191182 0.017

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