Confidence Intervals for Normal and T-Distributions

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 95% confidence.

• 1.699

• 0.56

• 0.537

• 19.325

Why do we use a t-distribution instead of a z-distribution for means? 

• It reflects greater variability present in small samples.
• It allows us to decrease the margin of error without increasing our confidence level.
• It allows us to use the population standard deviation.
• None of these.

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.

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. 

• (1070, 2530)
• (1273, 2326)
• (1799, 1801)
• None of these.

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

• 1.714
• 1.711
• 2.064
• None of these.

A research group wishes to estimate the mean amount of time (in hours) that members of a fitness center spend exercising each week. They want to estimate the mean within a margin of error of 0.5 hours with a 95% level of confidence. Previous data suggests that the standard deviation of the population is 2.2. Which of the following is the smallest sample size they could use?

• 60
• 75
• 90
• 180

A 90% confidence interval for a population mean is determined to be 800 to 900. If the confidence is increased to 95% confidence while the sample statistics and sample size remain the same, the confidence interval

• becomes wider
• becomes narrower
• does not change

A random sample of 100 visitors to a popular theme park spent an average of $142 on the trip with a standard deviation of $47.5. 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)
• ($132.77, $151.43)
• ($132.69, $151.31)
• ($140.88, $143.12)

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 sample size is too small for 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 batch is unknown.

A random sample of 100 Woodrow students found that 53 of them were in possession of a pencil. Create a 90% confidence interval for this sample.

• (0.45,0.61)
• (0.43,0.63)
• (0.53,0.63)