Erythromycin is a drug that has been proposed to possibly lower the risk of prem

Question

Erythromycin is a drug that has been proposed to possibly lower the risk of premature delivery. A related area of interest is its association with the incidence of side effects during pregnancy. Assume that 30% of all pregnant women complain of nausea between the 24th and 28th week of pregnancy. Furthermore, suppose that of 195 women who are taking erythromycin regularly during this period, 68 complain of nausea. Find the p-value for testing the hypothesis that incidence rate of nausea for the erythromycin group is greater than for a typical pregnant woman.

Explanation / Answer

In this problem we can test the incidence rate of nausea for the erythromycin group is greater than for a typical pregnant woman that means we can cheak the hypothes,

Ho: incidence rate of nausea for the erythromycin group is 30 % i.e. p=0.3

v/s

H1: incidence rate of nausea for the erythromycin group is gretter than 30 % i.e. p>0.3

from the given imformation , we clear that value of incident rate in whole population which is 30% i.e. population perportion is p=0.3

afrer getting survey we observed that 195 women who are taking erythromycin regularly during this period, 68 complain of nausea that means sample perportion p^=68/195=0.3487179.

for testing above hypothesis ,the test statistics as follows,

z=(p^-po)/sqrt(po(1-po)/n)

where p^=0.3487179. po=0.3 n=195   

z=(0.3487179-0.3)/sqrt(0.3*0.7/195)

z =1.484555

and p-value=p(z>1.48455)

=0.06883087

here z<1.64 and p -value is 0.06883087 which is gretter than 0.05. means we can weakly accept Ho.

i.e.  incidence rate of nausea for the erythromycin group is 30 % i.e. p=0.3.

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