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1. On February 8, 2002, the Gallup Organization released the results of a poll concerning American attitudes toward the 19th Winter Olympics in Salt Lake City, Utah. The poll results were based on telephone interviews with a randomly selected national sample of 1011 adults, 18 years and older, conducted February 4 - 6, 2002.

a) Suppose we wish to use the poll's results to justify the claim that more than 30 percent of Americans (18 years or older) say that figure skating is their favorite Winter Olympic event. The poll actually found that 32 percent of respondents reported that figure skating was their favorite event. If, for the sake of argument, we assume that 30 percent of Americans (18 years or older) say figure skating is their favorite event, calculate the probability of observing a sample proportion of 32 percent or more.

b) Based on the probability you computed in a, would you conclude that more than 30 percent of Americans (18 years or older) say that figure skating is their favorite Winter Olympic event? Explain.

2. Megacorp wants to determine whether the mean of all one-day travel expenses in Moscow exceeds \$500. A simple random sample of 35 one-day travel expenses is obtained from Megacorp travel expense files. The average one-day Moscow expense was found to be \$538 with a standard deviation of \$41.

a) Calculate the probability of observing a sample mean of at least \$538 if the actual population mean is \$500.

b) Based on you answer to a, do you think the mean of all one-day travel expenses for Megacorp employees in Moscow exceeds \$500? Explain.

3. Suppose that, for the sample size of n=100 measurements, we find a sample mean of 50 and a standard deviation of 2. Calculate confidence intervals for the population mean with the following confidence levels:

a) 95%

b) 82%

https://brainmass.com/statistics/probability/polling-8072

#### Solution Preview

Hi,
<br>Here the sequence of solutions is: 3,2,1. Hope you will have no problem.
<br>
<br>3.
<br>a.)
<br>given:
<br>n =100
<br><x> = 50
<br>SD = 2
<br>mu =? (mean)
<br>1-a = 95/100
<br>i.e.,
<br>P[<x> - t(a/2)*SD/sqrt(n) < mu < <x> + t(a/2)*SD/sqrt(n)] = .95
<br>Sol.:
<br>
<br>t(a/2) = 1.662 (from table)
<br>
<br>therefore,
<br>50 - 1.662 *2/sqrt(100) < mu < 50 + 1.662 * 2/sqrt(100)
<br>=> 49.6676 < mu < 50.3324 ...

#### Solution Summary

1. On February 8, 2002, the Gallup Organization released the results of a poll concerning American attitudes toward the 19th Winter Olympics in Salt Lake City, Utah. The poll results were based on telephone interviews with a randomly selected national sample of 1011 adults, 18 years and older, conducted February 4 - 6, 2002.

a) Suppose we wish to use the poll's results to justify the claim that more than 30 percent of Americans (18 years or older) say that figure skating is their favorite Winter Olympic event. The poll actually found that 32 percent of respondents reported that figure skating was their favorite event. If, for the sake of argument, we assume that 30 percent of Americans (18 years or older) say figure skating is their favorite event, calculate the probability of observing a sample proportion of 32 percent or more.

b) Based on the probability you computed in a, would you conclude that more than 30 percent of Americans (18 years or older) say that figure skating is their favorite Winter Olympic event? Explain.

2. Megacorp wants to determine whether the mean of all one-day travel expenses in Moscow exceeds \$500. A simple random sample of 35 one-day travel expenses is obtained from Megacorp travel expense files. The average one-day Moscow expense was found to be \$538 with a standard deviation of \$41.

a) Calculate the probability of observing a sample mean of at least \$538 if the actual population mean is \$500.

b) Based on you answer to a, do you think the mean of all one-day travel expenses for Megacorp employees in Moscow exceeds \$500? Explain.

3. Suppose that, for the sample size of n=100 measurements, we find a sample mean of 50 and a standard deviation of 2. Calculate confidence intervals for the population mean with the following confidence levels:

a) 95%

b) 82%

\$2.49