AP Statistics T Intervals

A quality control specialist at a glass factory must estimate the mean clarity rating for a new batch of glass using a sample of 18 glass sheets from the batch. Past investigations show that clarity ratings are normally distributed. The specialist decides to use a t-distribution rather than a z-distribution because ...

• The t distribution is more accurate than a z distribution.

• Clarity ratings for the entire bacth are normally distributed.

• The t-distribution will create a narrower interval than z.

• The standard deviation for the population is unknown.

A Gallup poll asked a random sample of U.S. adults, "Would you like to lose weight?" Based on this poll, the 95% confidence interval for the proportion of all U.S. adults who want to lose weight is 0.56 to 0.62. Interpret the confidence level.

• If Gallup took many random samples of U.S. adults and constructed a 95% confidence interval for each sample, about 95% of these intervals would capture the true proportion of all U.S. adults who want to lose weight.

• If Gallup took many random samples of U.S. adults and constructed a 95% confidence interval for each sample, about 95% of these intervals would not capture the true proportion of all U.S. adults who want to lose weight.

• If Gallup took many random samples of U.S. adults and constructed a 99% confidence interval for each sample, about 99% of these intervals would capture the true proportion of all U.S. adults who want to lose weight.

• If Gallup took many random samples of U.S. adults and constructed a 90% confidence interval for each sample, about 90% of these intervals would not capture the true proportion of all U.S. adults who want to lose weight.

The lifetime of a certain type of batter is known to be normally distributed with a standard deviation of 20 hours. A sample of 50 batteries had a mean lifetime of 120.1 hours. It is desired to construct a 99% confidence interval for the mean lifetime of this type of battery. What is the Confidence Interval to one decimal place?

• (114.5, 125.6)

• (112.8, 127.4)

• (112.82, 127.38)

• (114.56, 125.64)

Which of the following changes to a study would result in a narrower confidence interval?

• increasing the confidence level, increasing the sample size

• decreasing the confidence level, decreasing the sample size

• increasing the confidence level, decreasing the sample size

• decreasing the confidence level, increasing the sample size.

A study of a local high school with 2,000 students tried to determine the mean number of text messages that each student sent per day. The study surveyed a random sample of 253 students in the high school and found a mean of 186 messages sent per day with a standard deviation of 65 messages.

Would you use a one sample z interval or a one sample t interval for this situation?

• One Sample z Interval

• One Sample t Interval

One reason for using a t distribution instead of the standard Normal curve to find critical values when calculating a level C confidence interval for a population mean is that

• z can be used only for large samples

• z requires that you know the population standard deviation

• z requires that you can regard your data as an SRS from the population

• z requires that the sample size is at most 10% of the population size

• a z critical value will lead to a wider interval than a t critical value

Which of the following does NOT have an affect on the width of a confidence interval?

• Sample mean

• t*

• Sample standard deviation

• Sample size

Find the t* value for a 99% confidence interval with a sample of 23 people.

• 1.714

• 1.711

• 2.064

• None of these.

An IQ test was given to a simple random sample of 48 students at a certain college. The sample mean score was 105.2. Scores on this test are known to have a standard deviation of 10. It is desired to construct a 90% confidence interval for the mean IQ scores of the students at the college. Find the critical value.

• 0.90

• 1.984

• 1.645

• 1.676

The football coach randomly selected ten players and timed how long each player took to perform a certain drill. The times (in minutes) were: 13.2, 5.1, 7.5, 8.0, 12.7, 7.6, 13.8, 14.5, 7.7, and 10.5. Determine a 95 percent confidence interval for the mean time for all players.

• (2.30, 7.82)

• (7.82, 12.30)

• (7.72, 12.40)

• (2.40, 7.72)