Question 619963: Hi, I've been stuck on this question all day! Hoping I could get some help:
Whats the height of college hockey players? The avg height has been 68.3 inches. A random
sample of 14 hockey players gave a mean height of 69.1 inches. We may assume that x has a normal distribution with std dev of 0.9 inch. Does this indicate that the population mean height is
different from 68.3 inches? Use 5% level of significance
and this one as well please:
A random sample of 83 investment portfolios managed by Kendra showed that 62 of them met the targeted annual percent growth. A random sample of 112 portfolios managed by Lisa showed that 87 met the targeted annual percent growth. Find a 99% confidence interval for the difference in the proportion of the portfolios meeting target goals managed by Kendra compared with those managed by Lisa. Is there a difference in the proportions at the 99% confidence level?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! What is the height of college hockey players? The avg height has been 68.3 inches.
A random sample of 14 hockey players gave a mean height of 69.1 inches. We may assume that x has a normal distribution with std dev of 0.9 inch. Does this indicate that the population mean height is
different from 68.3 inches? Use 5% level of significance
Ho: u = 68.3 (claim)
Ha: u # 68.3
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t(69.1) = (69.1-68.3)/[0.9/sqrt(14)] = 0.2138
p-value = 2*p(0.2138< = t <= 100 when df = 13) = 2*tcdp(0.2138,100,13)
= 0.4170
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Since the p-value is greater than 5%, fail to reject Ho.
The mean height is 68.3.
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A random sample of 83 investment portfolios managed by Kendra showed that 62 of them met the targeted annual percent growth. A random sample of 112 portfolios managed by Lisa showed that 87 met the targeted annual percent growth.
--------
Find a 99% confidence interval for the difference in the proportion of the portfolios meeting target goals managed by Kendra compared with those managed by Lisa.
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Is there a difference in the proportions at the 99% confidence level?
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Kendra DATA:
sample proportion = phat2 = 62/83 = 0.75
Lisa DATA:
sample proportion = phat1 = 87/112 = 0.78
------
Ho: p1 - p2 = 0
Ha: p1 - p2 # 0
---------------------------------
std of proportion differences: sqrt[(p1q1/n1) + (p2q2/n2)]
= sqrt[(.75*0.25/83)+(0.78*0.22/112)] = 0.0616
---------------------------------------------------
z(difference) = (0.78-0.75)/0.0616 = 0.4872
----
sample proportion difference = 0.03
std of sample difference: sqrt[(0.75*0.25/83)+(0.78*0.22/112)] = 0.0616
-----
Margin of error: 2.5758*0.0616 = 0.1586
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99% CI: 0.03-0.1586 < p1-p2 < 0.03+0.1586
(-0.1286 , +0.1886)
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Conclusion: Since zero is in the 99% CI, fail to reject Ho.
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Cheers,
Stan H.
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