SOLUTION: Please help me with these as I am desperate....I am trying to teach myself statistics and it is not working. 2. Given a level of confidence of 95% and a population standard dev

Click here to see ALL problems on Probability-and-statistics

Question 406752: Please help me with these as I am desperate....I am trying to teach myself statistics and it is not working.
2. Given a level of confidence of 95% and a population standard deviation of 6, answer the following:
(A) What other information is necessary to find the sample size (n)?
(B) Find the Maximum Error of Estimate (E) if n = 84. Show all work.

3. A sample of 134 golfers showed that their average score on a particular golf course was 93.23 with a standard deviation of 6.25.
Answer each of the following (show all work
and state the final answer to at least two decimal places.):
(A) Find the 98% confidence interval of the mean score for all 134 golfers.
(B) Find the 98% confidence interval of the mean score for all golfers if this is a sample of 110 golfers instead of a sample of 134.
(C) Which confidence interval is larger and why? (Points : 6)
5. The diameters of grapefruits in a certain orchard are normally distributed with a mean of 5.91 inches and a standard deviation of 0.52 inches. Show all work.
(A) What percentage of the grapefruits in this orchard is larger than 5.86 inches?
(B) A random sample of 100 grapefruits is gathered and the mean diameter is calculated. What is the probability that the sample mean is greater than 5.86 inches?

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)

Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website!

Hi
Understand completely, recall there was a time I was self taught on most
ALL early math at any rate: (country school:)
Might recommend checking out the statistic's section on this site:
great learning tool selecting previous questions answered, of interest to You.
2. Given a level of confidence of 95% and a population standard deviation of 6, answer the following:
(A) What other information is necessary to find the sample size (n)?
confidence interval or 'margin of error'
(B) Find the Maximum Error of Estimate (E) if n = 84. Show all work.
ME = 1.96[6/sqrt(84)]
3. A sample of 134 golfers showed that their average score on a particular golf course was 93.23
with a standard deviation of 6.25.
Answer each of the following (show all work
and state the final answer to at least two decimal places.):
(A) Find the 98% confidence interval of the mean score for all 134 golfers.
ME = 2.3263[6.25/sqrt(134)]
CI: 93.23- ME < u < 93.23 + ME
(B) Find the 98% confidence interval of the mean score for all golfers if this is a sample of 110 golfers
instead of a sample of 134.
ME = 2.3263[6.25/sqrt(110)]
CI: 93.23- ME < u < 93.23 + ME
(C) Which confidence interval is larger and why? (Points : 6)
Note: the Larger the sample size, the smaller ME is and therefore the
confidence interval is larger with the smaller sample size.
5. The diameters of grapefruits in a certain orchard are normally distributed
with a mean of 5.91 inches and a standard deviation of 0.52 inches. Show all work.
(A) What percentage of the grapefruits in this orchard is larger than 5.86 inches?
z = 5.86-5.91 /.52 = -.05/.52 = -.0962
P(z > -.0962) = 1 - .4617 = .5383 53.83%
(B) A random sample of 100 grapefruits is gathered and the mean diameter is calculated.
What is the probability that the sample mean is greater than 5.86 inches?
z = -.05/ .52/sqrt(100) = -.05 /.052 = .9615
P(z > -.9615) = 1 -.16815 = .83185 83.185%
Folowing is a summary of Levels of Confidence & their critical regions
a a/2 crtical regions
90% 10 5% z <-1.645 z >+1.645
95% 5 2.50% z <-1.96 z >+1.96
98% 2 1% z <-2.33 z >+2.33
99% 1 0.50% z<-2.575 z >+2.575