More practice problems on confidence intervals: chapters 19 and 23


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.