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October 1, 2014

October 1, 2014

Posted by **Steven** on Tuesday, December 17, 2013 at 5:23pm.

• A researcher wants to determine the difference between the gestational period (period of pregnancy) for two groups of women, namely, those who had normal pregnancies (group A) and those who suffered from a certain complication during pregnancy known as preeclampsia (group B). The following data shows the gestational periods (in weeks) for a sample of 14 women from group A and of 13 women from group B.

Gestational periods for sample from group A: 40, 41, 38, 40, 40, 39, 39, 41, 41, 40, 40, 40, 39, 38

Gestational periods for sample from group B: 38, 30, 32, 42, 30, 38, 35, 32, 38, 39, 29, 29, 32

Find a 95% confidence interval for the difference between the gestational periods of group A and group B, i.e., group A mean minus group B mean. Assume that the variances of the two groups are equal. State any other assumptions you make.

Hint: This relates to confidence interval for

- statistics 3. -
**MathGuru**, Tuesday, December 17, 2013 at 6:26pmHere are a few ideas to get you started.

First problem:

Use a formula to find sample size.

Here is one for proportions:

n = [(z-value)^2 * p * q]/E^2

... where n = sample size, z-value is 1.645 (90% confidence interval), p = .5 (when no value is stated), q = 1 - p, ^2 means squared, * means to multiply, and E = 0.1 listed in the problem.

Plug values into the formula and calculate n.

Second problem:

You will need to use a two-sample confidence interval formula. You will also need to calculate mean and standard deviation for both groups.

Here is one:

CI95 = (x1 − x2) ± z-value * √(s1^2/n1 + s2^2/n2)

(Note: * means to multiply)

x1 = mean A

x2 = mean B

z-value = 1.96

s1^2 = standard deviation A (squared)

s2^2 = standard deviation B (squared)

n1 = sample size A

n2 = sample size B

Plug the values into the formula and calculate the interval.

I'll let you take it from here.

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