Question 407275: I am taking an online statistics course and really need help as I have no teacher to talk to. Can you please help me answer these:
1. Consider a population with µ = 99.4 and ð = 5.55 (Points : 6)
(A) Calculate the z-score for x– = 97.3 from a sample of size 38.
(B) Could this z-score be used in calculating probabilities using Table 3 in Appendix B of the text? Why or why not?
4. Assume that the population of heights of female college students is approximately normally distributed with mean of 64.64 inches and standard deviation of 6.02 inches. A random sample of 98 heights is obtained. Show all work.
(A) Find the mean and standard error of the x distribution
(B) Find P( -
x > 63.75)
6. A researcher is interested in estimating the noise levels in decibels at area urban hospitals. She wants to be 98% confident that her estimate is correct. If the standard deviation is 5.09, how large a sample is needed to get the desired information to be accurate within 0.57 decibels? Show all work. (Points : 6)
This is the answer I got....is it right?
A = 1 – 0.98 = .02
0.02/2 = 0.01
0.57 - .01 = 0.56 finding the closest on the z table 3 = 5
Z A/2 = 5
E = 0.57 and o = 5.09
N = [(5)(5.09)/0.57]2 = [25.45/0.57]2 = 44.62 = 1989.16
So a sample of 1989 is needed to get the desired info to be accurate within 0.57 decibels
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1. Consider a population with µ = 99.4 and ð = 5.55 (Points : 6)
(A) Calculate the z-score for x– = 97.3 from a sample of size 38.
z(97.3) = (97.3-99.4)/[5.55/sqrt(38)] = -2.3325
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(B) Could this z-score be used in calculating probabilities using Table 3 in Appendix B of the text? Why or why not?
If that Appendix is a z-chart the answer is yes.
If it is a t-chart that could also be used.
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4. Assume that the population of heights of female college students is approximately normally distributed with mean of 64.64 inches and standard deviation of 6.02 inches. A random sample of 98 heights is obtained. Show all work.
(A) Find the mean and standard error of the x distribution
Your problem statement gives the mean and std.
Maybe you are looking for the mean and std of the
distribution of sample means which would be
mean of x-bars = 64.64 and std of the x-bars = 6.02/sqrt(98)
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(B) Find P(x > 63.75)
z or t of 63.75 = (63.75-64.64)/6.02 = -0.1478
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6. A researcher is interested in estimating the noise levels in decibels at area urban hospitals. She wants to be 98% confident that her estimate is correct. If the standard deviation is 5.09, how large a sample is needed to get the desired information to be accurate within 0.57 decibels? Show all work.
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n = [z*s/E]^2*pq
n = [2.3263*5.09/0.57]^2*(1/2)(1/2)
n = 431.55*(1/4)
n = 107.88
Rounding up you get n = 108
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Cheers,
Stan H.
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