In 1998 a San Diego Reproductive clinic reported 49 live births to 207 women under the age of 40 who had previously been unable to get pregnant.
Using the 90% confidence interval calculated above, would it be misleading for the clinic to report a success rate of 25%?
A) Yes, because the 90% confidence interval extends above 25%
B) Yes, because we should have calculated a 99% confidence interval to be more certain
C) No, because the 90% confidence interval includes 25%
D) Yes, because the 90% confidence includes 25%; to advertise this success rate the lower bound of the 90% confidence interval should above 25%
Wouldn't the answer be C since the interval falls between 0.19-0.29?
Ans C
Includes 25%
[.207, .266]
No, the answer would not be C. Let me explain why.
In this scenario, the clinic reported 49 live births out of 207 women, which indicates a success rate of 49/207 = 0.236 or 23.6%. The question asks whether it would be misleading for the clinic to report a success rate of 25%.
To determine if it would be misleading, we need to consider the 90% confidence interval that was previously calculated. The 90% confidence interval was determined to be 0.19 to 0.29, which means that we can be 90% confident that the true success rate falls within this range.
Now, let's compare the reported success rate of 25% to this confidence interval. The reported success rate of 25% is outside the range of the confidence interval (0.19 to 0.29). Therefore, it would be misleading for the clinic to report a success rate of 25% because it does not fall within the 90% confidence interval.
Therefore, the correct answer is A) Yes, because the 90% confidence interval extends above 25%.