Confidence Intervals

Which z-value is used for a 95% confidence interval?

• 1.65
• 1.96
• 2.58
• 2.33

Tags

CCSS.HSS.ID.A.4

Given the same mean, SD, and sample size, a confidence interval of 80% would be wider than a confidence interval of 90%

• True
• False

Tags

CCSS.HSS.IC.B.4

CCSS.HSS.IC.A.1

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 sample standard deviation of $950. Construct a 99% confidence interval for the repair cost. 

• (1069.77, 2530.23)
• (1077.13, 2522.87)
• (1167.15, 2432.85)
• None of these.

A sample of 40 people owned an average of 1.35 apple products. It is known from a previous study that the population standard deviation of apple product ownership is 0.63 products. Construct a 95% confidence interval.

• (1.11, 1.58)
• (1.15, 1.55)
• (1.19, 1.51)
• (1.09, 1.61)

Tags

CCSS.HSS.IC.B.4

CCSS.HSS.IC.A.1

CCSS.HSS.ID.A.4

A sample of 30 women finds the mean height to be 62.3 in and the standard deviation to be 1.5 in. Find the margin of error of women's heights with a 95% CL.

• 1.699
• 0.56
• 0.537
• 19.325

Tags

CCSS.HSS.IC.B.4

A 90% confidence interval for the average salary of all CEOs in the electronics industry was constructed using the results of a random survey of 45 CEOs. The interval was ($139,048, $154,144). Give a practical interpretation of the interval.

• 90% of the sampled CEOs have salaries that fell in the interval $139,048 to $154,144
• We are 90% confident that the mean salary of all CEOs in the electronics industry falls in the interval $139,048 to $154,144. 
• There is a 90% chance that CEOs in the electronics industry have salaries that fall between $139,048 to $154,144
• We are 90% confident that the mean salary of the sampled CEOs falls in the interval $139,048 to $154,144.

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

CCSS.HSS.IC.A.1

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 μ in test scores for all sixth-graders in the city taking this course is found to be (12.6, 14.8). Which one of the following is a correct interpretation of this confidence interval?

• Ninety percent of the 85 sixth-graders in the sample improved their reading comprehension scores by 12.6 to 14.8 points.
• The probability is 0.90 that the true mean improvement in reading comprehension is between 12.6 and 14.8.
• We are 90% confident that the true improvement in reading comprehension scores is captured by the interval 12.6 to 14.8.

Agricultural researchers plant 100 plots with a new variety of corn and measure the mean yield for these plots in bushels per acre. They treat the 100 plots as a simple random sample of possible plots of corn (as researchers often do) and report a 95% confidence interval for the mean corn yield of (128.4, 131.6) bushels per acre. Which of the following is a correct interpretation of the 95% confidence level?

• We are 95% confident that the true mean yield for this new variety of corn is captured by the interval (128.4, 131.6) bushels per acre
• If many intervals were constructed this way from many independent sets of 100 plots, 95% of the intervals would capture the true mean corn yield.
• If many intervals were constructed this way from many independent sets of 100 plots, 95% of the time the true mean corn yield would be in the interval (128.4, 131.6) bushels per acre.

Tags

CCSS.HSS.IC.B.4

A radio talk show host is interested in the proportion p of adults in his listening area who think that the drinking age should be lowered to 18. To find this proportion, he poses the following question to his listeners: “Do you think that the drinking age should be reduced to 18?” He asks listeners to phone in and vote “yes” or “no” depending upon their opinions. Of 200 people who phone in, 140 answer “yes.” The standard error of the sample proportion  of “yes” votes among those who phone in is

• .21
• .032
• .00105