From a large population of married couples, researchers selected a random sample of four hundred couples.
The heights of the married men are approximately normally distributed, with a mean of 70 inches and a standard deviation of 3.
The heights of the married women are approximately normally distributed, with a mean of 65 inches and a standard deviation of 2.5.
There were twenty couples in which the wife was taller than her husband; in the other 380 couples, the husband was taller than his wife.
Find a 95% confidence interval for the proportion of the population for which the wife is taller than her husband.
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95% conf. int. for prop. of pop. w/ wife taller than hubby 
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95% conf. int. for prop. of pop. w/ wife taller than hubby
by maggiemagnet on Wed Jun 24, 2009 4:21 pm
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maggiemagnet - Posts: 250
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by stapel_eliz on Wed Jun 24, 2009 6:46 pm
I'm not good at this stuff, but let's give it a whirl.
The proportion is 20/400 = 2/40 = 1/20 = 0.05, so p = 0.05 and q = 0.95.
The z-score for 0.95 is 1.96 (or thereabouts; your book may use a different number of decimal places). So the margin of error is:
. . . . .\sqrt{\frac{0.0475}{400}}\, \approx\, 0.0213586048...)
...or about 0.0214. Then the interval should be 0.05 +/- 0.0214.
The proportion is 20/400 = 2/40 = 1/20 = 0.05, so p = 0.05 and q = 0.95.
The z-score for 0.95 is 1.96 (or thereabouts; your book may use a different number of decimal places). So the margin of error is:
. . . . .
...or about 0.0214. Then the interval should be 0.05 +/- 0.0214.
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stapel_eliz - Posts: 1706
- Joined: Mon Dec 08, 2008 4:22 pm
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