A clinical trial tests a method designed to increase the probability of conceivi

Question

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 380 babies were born, and 323 of them were girls. Use the sample data to construct a 99% confidence interval estimate of the percentage of girls born. Based on the result, does the method appear to be effective?

___< p <___ (Round to three decimal places as needed.)

Does the method appear to be effective?

No, the proportion of girls is not significantly different from 0.5.

Yes, the proportion of girls is significantly different from 0.5.

Explanation / Answer

Solution:-
a) given a 1-0.99 = 0.01
the critical value is Z(0.005) = 2.58
(from the standard normal table)

=> p = 323/380 = 0.85, q = 1-p = 0.15
  
So 99% confidence interval is
p +/- Z*sqrt(pq/n)
--> 0.85 +/- 2.58*sqrt((0.85*0.15)/380)
--> 0.85 +/- 2.58*sqrt(0.00034)
--> 0.85 +/- 2.58*0.0184
--> 0.85 +/- 0.0475
--> (0.803,0.898) Rounded

--->0.803< p <0.898

===> option a. No, the proportion of girls is not significantly different from 0.5.

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