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 5 of them smoked 21 or more cigarettes during pregnancy. Test the​ researcher's statement at the α=0.05 level of significance. Identify the correct null and alternative hypotheses.

H0​: p ()0.05

H1: p() 0.05

Use technology to find the​ P-value. P-value= ​(Round to three decimal places as​ needed.)

Is there sufficient evidence to support the​ obstetrician's statement? A.No​, because the​ P-value is greater than α there is not sufficient evidence to conclude that the percentage of mothers who smoke 21 or more cigarettes during pregnancy is less than 5​%, meaning we do not reject the null hypothesis. B.Yes​, because the​ P-value is greater than α there is sufficient evidence to conclude that the percentage of mothers who smoke 21 or more cigarettes during pregnancy is less than 5​%, meaning we do not reject the null hypothesis. C.No​, because the​ P-value is less than α there is not sufficient evidence to conclude that the percentage of mothers who smoke 21 or more cigarettes during pregnancy is less than 5​%, meaning we reject the null hypothesis. D. Yes​, because the​ P-value is less than α there is sufficient evidence to conclude that the percentage of mothers who smoke 21 or more cigarettes during pregnancy is less than 5​%, meaning we reject the null hypothesis.

Answer 1

Givwn that previously, 5​%( p = 0.05)  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 n = 130 pregnant mothers and finds that X = 5 of them smoked 21 or more cigarettes during pregnancy.

based on the claim the hypotheses are:

Ho : p = 0.05

Ha : p < 0.05

Now based on the hypothesis it will be left-tailed test.

Rejection region:

At 0.05 level of significance reject the Ho if P-value is less than 0.05.

Sample proportion:

The sample proportion is calculated as:

Test statistic:

P-value:

The P-value is computed using the excel formula for Z-distribution which is =NORM.S.DIST( -0.604, TRUE), thus the P-value is computed as:

Conclusion:

Since the P-value is greater than 0.05 hence we fail to reject the null hypothesis hence the conclusion is:

A.No​, because the​ P-value is greater than α there is not sufficient evidence to conclude that the percentage of mothers who smoke 21 or more cigarettes during pregnancy is less than 5​%, meaning we do not reject the null hypothesis.