SOLUTION: Error finding the Margin of Error Q: A study was conducted to determine the proportion of people who dream in black and white instead of color. Among 400 people over the age of

Question 1069921: Error finding the Margin of Error
Q: A study was conducted to determine the proportion of people who dream in black and white instead of color. Among 400 people over the age of 55, 121 dream in black and white, and among 380 people under the age of 25, 56 dream in black and white. Use a 0.05 significance level to test the claim that the proportion of people over 55 who dream in black and white is greater than the proportion for those under 25.
I calculated the following:
p-hat1= 0.3025
p-hat2 = 0.1474
z = 5.1704
p-value = 0
What is the conclusion based on the hypothesis test?
The P-value is less than the significance level of alpha=0.05, so reject the null hypothesis. There is
sufficient evidence to support the claim that the proportion of people over 55 who dream in black and white is greater than the proportion for those under 25.
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Test the claim by constructing an appropriate confidence interval.
The 90% confidence interval is [ ] < p 1 - p 2 < [ ]
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I can't figure out how to calculate the margin of error. I have the equation (E = z a/2, etc), but I get answers that aren't correct. I don't need the answers because there will be about 20 of them. I need help figuring out the process because I'm doing something wrong.
Thank you in advance.
~J
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
ME = z*sqrt[(p1q1/n1 + p2q2/n2)]
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Also,
Since the 95% CI is (0.098,0.213), 2*ME = 0.213-0.098 = 0.115
ME = 0.115/2 = 0.575
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Cheers,
Stan H.
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