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# Statistics

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1. Suppose you have drawn a simple random sample of 400 students from University X and recorded how much money each student spent on text books in Fall 2009. For your sample, sample mean (X-bar) is \$500, and sample standard deviation (s) is \$100. Construct a 99% confidence interval for the mean.

2. Suppose you have drawn a simple random sample of 400 students from University X. In this sample, 100 students own Blackberries. Let p denote the proportion of University X students who own Blackberries. Construct a 99% confidence interval for p.

3. Suppose you want to use a simple random sample to determine, within ± \$20 and at a 99% level of confidence, the average amount of money a Syracuse University undergraduate student spends on text books in a year. If you know that the standard deviation is 100, compute the sample size you need.

4. Let m denote the average amount of money a household in County X spent on automobile insurance in 2009. You want to use a simple random sample to determine m within ±0.06 m at a 99% level of confidence. You have no prior idea of m. However, you expect that the amount a household spends on automobile insurance is roughly proportional to the annual income of the household. Also, from secondary data, you know that the mean annual income of County X was \$75,000, and the standard deviation in annual income in County X was \$30,000. Compute the sample size you need.

5. There are 150,000 full time college students in State X. You want to use a simple random sample to determine, within ± 6000 and with 99% confidence, how many of these 150,000 students own Blackberries. If you have no prior idea of how many, determine the sample size you need.

6. Suppose you want to study the population of full-time college students in State X. In this state, there are 150,000 full-time college students. Out of these students, 30,000 students study at private schools, and the other 120,000 students study at state schools. You have drawn a stratified sample using private school students and state school students as the two strata. For each student selected, you recorded how much money the student spent on tuition and other college expenses (expense) in 2009. The results are as follows:

(1) Private School Students: n1 = 100, X-bar-1 = 30,000, s1 = 15, 000.

(2) State School Students: n2= 400, X-bar-2 = 15,000, s2= 6000.

6. (a) Construct a 99% confidence interval for the total expense by all the 30,000 private school students, combined, in 2009.
6. (b) Let m denote the mean expense of a full-time college student in State X in 2009. Construct a 99% confidence interval for m.

7. In State X, there are 150,000 full-time college students. Out of these students, 30,000 students study at private schools, and the other 120,000 students study at state schools. You have drawn a stratified sample using private school students and state school students as the two strata and, for each student, recorded if the student owns an Apple computer.

Results:
(1) Private School Students: n1 = 100. In this sample of 100, 30 students own Apple computers.
(2) State School Students: n2 = 400. In this sample of 400, 80 students own Apple computers.

Let p denote the proportion of the overall full-time college student population in State X that own Apple computers. Construct a 99% confidence interval for p.

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#### Solution Preview

Dear student, please refer to the attachment for the solutions. Thank You.

Sampling problems

1.Suppose you have drawn a simple random sample of 400 students from University X and recorded how much money each student spent on text books in Fall 2009. For your sample, sample mean (X-bar) is \$500, and sample standard deviation (s) is \$100. Construct a 99% confidence interval for μ.

Solution:
The confidence interval for the mean is given as:

(x ̅-(z_(α/2)×s)/√n,x ̅+(z_(α/2)×s)/√n)

where,Sample Size,n=400
Sample Mean,x ̅=500
Sample Standard Deviation,s=100

z_(α/2) for 99% confidence=2.58
Thus the 99% confidence interval for the mean is given as:

(500-(2.58×100)/√400,500+(2.58×100)/√400)

=(500-12.9,500+12.9)

=(487.1,512.9)

2. Suppose you have drawn a simple random sample of 400 students from University X.
In this sample, 100 students own Blackberries. Let π denote the proportion of University X
students who own Blackberries. Construct a 99% confidence interval for π.

Solution:
The confidence interval for proportion is given as
p±z_((1-α/2) ) √(p(1-p)/n)
where,p represents the proportion of success=100/400=0.25,
n represents sample size=400
z_((α/2) ) for 99% confidence level=2.58
Therefore, the required Confidence Interval is given as
p±z_((α/2) ) √(p(1-p)/n)=0.25±2.58√((0.25(1-0.25))/400)=0.25±0.0559

3. Suppose you want to use a simple random sample to determine, within ± \$20 and at a 99% level of confidence, the average amount of money a Syracuse University undergraduate student spends on text books in a year. If you know that σ ≈ 100, compute the sample size you need.

Solution:
Sample Size, n is given as ...

#### Solution Summary

This solution is comprised of detailed explanation and step-by-step calculation of the given problems and provides students with a clear perspective of the underlying concepts.

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