MAT 247: SOLUTIONS
For each problem, first decide whether you are trying to find a confidence interval for the population proportion or population mean.† If itís the mean, decide whether you should use a z-interval or t-interval.† Finally, in each case, write a sentence summarizing the meaning of the interval.
1. A random sample of prices for large camping tents gave the following prices, in dollars: 115, 110, 140, 135, 80, 110, 150, 210, 250, 120, 230, 130.† Find and interpret a 90% confidence interval for the mean price of a tent.† Assume a normal population.
Use a t-interval.† (120.85, 175.82). We are 90% confident that the mean price of all large camping tents is between $120.85 and $175.82.
2. In a random sample of 250 teenagers aged 14 to 17, 87 reported using illegal drugs at least once (not including alcohol, which would of course be illegal for that age group).† Find a 92% confidence interval for the proportion of all teens in that age group who have used illegal drugs at least once?† What is the margin of error?† Is it a reasonable margin of error to report in a poll?† If not, what could be done to improve it?
(0.295, 0.400).† We are 92% confident that between 29.5% and 40% of all teens aged 14 to 17 have used illegal drugs.† The margin of error is 0.0525=5.25%, which is reasonable to report (a 3% margin of error is more standard).† To improve it we could obtain a larger sample size.
3. A large loan company specializes in making loans for used cars.† Based on historical data, they know that the standard deviation of all their loans has been $750.† A random sample of 225 loans had a mean of $8200.† Find and interpret a 95% confidence interval for the mean of all used car loans from that company.
Since the sample size is large and the population standard deviation is known, we can use a z-interval: ($8102, $8298)† We are 95% confident that the mean amount of all used car loans is between $8102 and $8298.
4. In the last problem, what was the margin of error for the estimate of the mean amount of loan?† How large would the sample have to be to have a margin of error of $50?
The margin of error was $98. If we wanted a margin of error of $50 at the same confidence level we would need a sample size of 865.
5. A statistician working for a news agency wants to estimate the proportion of all registered voters who believe that the economy is the most important election issue facing Americans today.† She wants to use a 95% confidence interval and must have no more than a 3% margin of error.† How large a sample is needed? (since she has no preliminary estimate for the proportion, use p = Ĺ in the formula.)
Use n = 1068.
6. Suppose the statistician in the last question is able to survey 1000 registered voters, and 596 of them say they believe the economy is the most important election issue.† Find a 95% confidence interval.† What is the margin of error?† Does this make sense in light of your answer to the last question?† Explain.
We are 95% confident that between 56.6% and 62.6% of all registered voters believes that the economy is the most important issue.† The margin of error is 0.0304.† It is slightly higher than 3% since we used a sample that was a bit smaller than the 1068 we would need for an exact margin of 3%.
7. How many calories are there in 3 ounces of French fries?† It depends on where you get them.† Good Cholesterol Bad Cholesterol, by Roth and Streicher, gives the data from eight popular fast-food restaurants.† The data (in calories) are
222††††††† 255††††††† 254††††††† 230††††††† 249††††††† 222††††††† 237††††††† 287
††††††††††††††† Use these data to find a 99% confidence interval for the mean calorie count in 3 ounces of French fries obtained from fast food restaurants.† Again, assume the population of all of these calorie counts is normal.
††††††††††††††† Use a t-distribution since the population standard deviation is not known.† We are 99% confident that the mean calorie count in a 3-ounce bag of fries is between 217.61 and 271.39 calories.