Question

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 conceive.

In 1998 a San Diego reproductive clinic reported 49 live births to 207 cnomen under the age oy been unabhe ta 6. conceive a. Find a 90% confidence interval for the success rate at this clinic.

Answer 1

6.

(a) = 49/207 = 0.237, 1 - = 0.763, n = 207, = 0.10

The Zcritical (2 tail) for $\alpha$ = 0.10, is 1.645

The Confidence Interval is given by ME, where

$ME = Zcritical * \sqrt{\frac{\hat{p}(1-\hat{p})}{n} }= 1.645 * \sqrt{\frac{0.237*0.763}{207}} = 0.0486$

The Lower Limit = 0.237 - 0486 = 0.1884 $\approx$ 0.188 ( in terms of percentage = 18.84%)

The Upper Limit = 0.237 + 0.0486 = 0.2856 $\approx$ 0.286 ( in terms of percentage =28.56%)

The Confidence Interval is (0.188 , 0.286)

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(b) 90% confidence level means that around 90% of confidence interval of size 207 will contain the true population proportion of women aged less than 40 and who were previously unable to conceive but have conceived in this clinic.

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(c) No, It would not be misleading to advertise a 25% success rate since the value of 25% lies within the limits of the confidence interval (0.188 and 0.286)

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(d) Sample size required when ME is half of the present value, all other conditions remaining the same.

The ME = Zcritical * SQRT(p * q/n).

Squaring and solving, n = (Zc/ME)² * p * q

Here ME = 0.0486/2 = 0.0243, p = 0.237, q = 0.763

The Zc at $\alpha$ = 0.10 is 1.645

Therefore n = (1.645/0.0243)² * 0.237 * 0.763 = 828.69

Therefore n = 829 (Rounding to the nearest Integer)

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(e) The concern here is that we are having to assume that this sample would be less that 10% of all such births in all the clinics.

Secondly There are only 40 such birth in the clinic and we have taken all of them into consideration, therefore this cannot be considered asa simple random sample.

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