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In 2001​, a reproductive clinic reported 43 live births to 169 women under the age of​...

In 2001​, a reproductive clinic reported 43 live births to 169 women under the age of​ 38, but only 9 live births for 75 clients aged 38 and older. Is there evidence of a difference in the effectiveness of the​ clinic's methods for older​ women? Determine the test statistic, the p-value, and if there is a difference, estimate the difference p1-p2 with a 95% confidence interval.

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SOLUTION:

From given data,

In 2001​, a reproductive clinic reported 43 live births to 169 women under the age of​ 38, but only 9 live births for 75 clients aged 38 and older. Is there evidence of a difference in the effectiveness of the​ clinic's methods for older​ women? Determine the test statistic, the p-value, and if there is a difference, estimate the difference p1-p2 with a 95% confidence interval.

= 43/169 = 0.254

n1 = 169

= 9/75 = 0.12

n2 = 75

= (n1 * + n2 * )/(n1 + n2)

= (169 * 0.254 + 75 * 0.12​​​​​​​ )/(169 +75)

= 51.926 / 244

= 0.21281

q = 1- = 1- 0.21281 = ​​​​​​​0.78719

SE = sqrt(p * q (1/n1 + 1/n2) )

SE = sqrt(0.21281 * 0.78719 (1/169 + 1/75) )

SE = 0.056788

Determine the test statistic

the test statistic = Z = ( - ) / SE

Z = (0.254 - 0.12​​​​​​​) / 0.056788

Z = 2.35

Determine the p-value

For two tailed test,

P(|Z| > 2.35) = area on left of -2.35 and on right of 2.35

P(|Z| > 2.35) = 2* P(Z < - 2.35)

P(|Z| > 2.35) = 2* 0.00939

P(|Z| > 2.35) = 0.01878

Estimate the difference p1-p2 with a 95% confidence interval.

95% confidence interval

Confidence interval is 95%

95% = 95/100 = 0.95

= 1 - Confidence interval = 1-0.95 = 0.05

/2 = 0.05 / 2

= 0.025

Z/2 = Z0.025 = 1.96

Confidence interval :

( - ) Z/2 * SE

(0.254 - 0.12​​​​​​​) 1.96 * 0.056788

0.134 0.11130448

(0.134 - 0.11130448 , 0.134 + 0.11130448)

(0.0226 , 0.2453)

0.0226 < p1-p2 < 0.2453

milcah answered 1 year ago
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