8.3.1 - Confidence Intervals for Means

In a crash test of 15 troopers, collision repair costs are found to have a distribution that is approximately normal, with a mean of $1800 and a standard deviation of $950. Construct a 99% confidence interval for the repair cost for the population.

• (1070, 2530)

• (1273, 2326)

• (1799, 1801)

• (1776, 1823)

Tags

CCSS.HSS.IC.B.4

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.

• 12.30 < μ < 7.82
•  7.82 < μ < 12.30
• 7.72 < μ < 12.40
• 12.40 < μ < 7.72

Tags

CCSS.HSS.IC.B.4

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.

Tags

CCSS.HSS.IC.B.4

What is the t value associated with 90% confidence and a sample size of 15?

• 1.341
• 1.345
• 1.761
• 1.753

Tags

CCSS.HSS.ID.A.4

A random sample of 100 visitors to a popular theme park spent an average of $142 on the trip with a sample standard deviation of $47.50. Which of the following would be the 98% confidence interval for the mean money spent by all visitors to this theme park?

• ($130.77, $153.23)
• ($130.96, $153.07)
• ($132.69, $151.31)
• ($140.88, $143.12)

Tags

CCSS.HSS.IC.B.4

Suppose that $51,759 was the median income for all households in 2012. Does the interval ($51,484, $52,394) provide convincing evidence that the median household income increased in 2013? 

• Yes
• No

Tags

CCSS.HSS.IC.B.4

A random sample of 85 sixth-graders in a large city take a course designed to improve scores on a reading comprehension test. Based on this sample, a 90% confidence interval for the mean improvement in test scores for all sixth-graders in the city taking this course is found to be (12.6, 14.8). Which of the following are the sample mean and margin of error on which this interval is based?

• Sample Mean = 13.7; Margin of Error = 1.1
• Sample mean = 13.7; Margin of Error = 2.2
• Sample mean is unknown; Margin of Error = 2.2

Tags

CCSS.HSS.IC.B.4

15 women were surveyed regarding the average number of books they read in a year. The sample mean was 42.07 with a standard deviation of 22.03. Construct a 90% confidence level for the average number of books women read in a year.

• (32.05, 52.09)

• (25.09, 34.14)

• (42.07, 52.07)

• (22.03, 65.75)

Tags

CCSS.HSS.IC.B.4

Choose the appropriate confidence interval for the situation:

A random sample of 30 movie theaters showed that the average price for a movie was $10 with a standard deviation of $1.20.

• z confidence interval for a mean

• t confidence interval for a mean

• z confidence interval for a proportion

• t confidence interval for a proportion

Tags

CCSS.HSS.ID.A.4

Compute a 90% confidence interval for the situation:

A random sample of 30 movie theaters showed that the average price for a movie was $10 with a standard deviation of $1.20.

• (9.71, 10.29)

• (9.934, 10.066)

• (9.64, 10.36)

• (9.628, 10.372)

Tags

CCSS.HSS.IC.B.4