STAT 411 Statistical MethodsI Prof. Hokwon Cho WorksheetName (Summer-11 2018) Di
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STAT 411 Statistical MethodsI Prof. Hokwon Cho WorksheetName (Summer-11 2018) Disc. Session: Chapter 9. Estimation of the Mean ? and Proportion p 1. A sample of 68 packages mailed from the McCarran Airport Post office showed a mean mailing charge of $14.85 with a standard deviation of $5.70. (a) Find the point estimate of u, and the margin of error associated with this estimate. (b) Construct a 99% confidence interval for the mean mailing charge for all packages mailed from this post office. Give a comment. 2. Suppose a sample of 16 packages selected from the University post office. The mean mailing charge, x was $12.70 with a standard deviation of $3.80. Assume that the mean mailing charges have a normal distribution. Construct a 95% confidence interval for the mean mailing charge for all packages mailed from this post office. Give an interpretation. 3. According to Time/CNN poll, 58% of phone interviews responded that they were happier than their parents. Suppose that this percentage is based on a random sample of 450 U.S. adults. (a) Construct a 95% confidence interval for the proportion of all U.S. adults who feel that they are happier than their parents (b) Find the sample size that would limit the maximum error to be within 0.025 of the population proportion for a 95% confidence interval. (c) In (b), what will be the sample size for the most conservative case? (i.e., this is the largest sample size) 4. In a certain manufacturing company, it is questioned that the resistance of wire A is greater than the resistance of wire B. An experiment on the wires (6 measurements each type) shows the following results (in 10'ohms): Wire B 1.35, 1.40, 1.36, 1.42, 1.38, 1.40 1.385 a) Assuming equal population variances, find the 90% confidence interval for the difference in the population means of two types of wires. (b) What conclusion do you draw from the result in (a)? Justify your answerExplanation / Answer
Soluion3:
given sample proportion=58%=0.58
n=sample size=450
need to find 95% confdence interval for population proportion,p
formula is
p^-Zsqrt(p^(1-p^)/n,p^+Zsqrt(p^(1-p^)/n
Z crit for 95% =1.96
therefore
0.58-1.96*sqrt(0.58*(1-0.58)/450),0.58+1.96*sqrt(0.58*(1-0.58)/450)
0.5344,0.6256
0.5344<p<0.6256
lower limit=0.5344
upper limit=0.6258
Solutionb:
E=0.025
n=Z^2*p^(1-p^)/E^2
=1.96^2*0.58*(1-0.58)/0.025^2
n=1497
required sample size=1497
Solutionc:
1497
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