Constructing Confidence Intervals

A sample of 400 racing cars showed that 80 cars cost over $700,000. What is the 99% confidence interval of the true proportion of cars costing over $700,000 if the margin of error is 0.052? (Hint, find the point estimate knowing that 80 of the 400 cars cost over $700,000). Remember CI = point estimate±margin of error

• (0.001, 0.005)

• (0.566, 0.693)

• (0.148, 0.252)

• (0.023, 0.045)

Answer explanation

Point estimate: 80/400 = 0.20, Margin of Error = 0.052

Lower bound: 0.2 - 0.052 = 0.148
Upper bound: 0.2 + 0.052 = 0.252

(0.148, 0.525)

Thirty randomly selected students took the calculus final. If the sample mean was 92 and the margin of error was 4.73, construct a 99 percent confidence interval for the mean score of all students from which the sample was gathered. Remember CI = point estimate ± margin of error

• 89.08 < μ < 94.92

• 87.27 < μ < 96.73

• 87.29 < μ < 96.71

• 87.77 < μ < 96.23

Answer explanation

point estimate = 92, margin of error = 4.73

Lower bound: 92 - 4.73 = 87.27
Upper bound: 92 + 4.73 = 96.73

 $87.27<\mu<96.73$  

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 if the margin of error is 730. Remember: CI = point estimate ± margin of error

• (1070, 2530)

• (1273, 2326)

• (1799, 1801)

• (1776, 1823)

Answer explanation

 $point\ estimate\ \pm m\arg in\ of\ error$  
Point estimate = 1800, Margin of error = 730

Lower bound: 1800 - 730 = 1070
Upper bound: 1800 + 730 = 2530

(1070, 2530)

A recent study of 750 Internet users in Europe found that 35% of Internet users were women. What is the 95% confidence interval of the true proportion of women in Europe who use the Internet if the margin of error is 0.034? Remember CI =  

• 0.321 < p < 0.379

• 0.316 < p < 0.384

• 0.309 < p < 0.391

• 0.305 < p < 0.395

Answer explanation

Point estimate = 0.35, Margin of error = 0.034.

Lower bound: 0.35 - 0.034 = 0.316

Upper bound: 0.35 + 0.034 = 0.384

0.315 < p < 0.384

The manufacturer of a smart watch claims that individuals who pay attention to how many steps they take per day will inadvertently take more steps per day than individuals who pay no attention to how many steps they take per day. To investigate this claim, the manufacturer conducts a study to estimate the difference in the mean number of steps taken by those that pay attention to how many steps they take per day and those that do not. To do so, 40 volunteers are recruited. Half of the volunteers are randomly assigned to receive a smart watch and are taught how to use it to track their steps. The other half of the volunteers are given a wristband to wear, but are not informed that the wristband is tracking their steps. The volunteers are monitored for 30 days. The mean and standard deviation of the number of steps taken per day are computed for each group. The data are shown above. Which of the following is a 99% confidence interval for the difference in the mean number of steps taken by all people like these that do and do not pay attention to the number of steps they take per day using df = 19?

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Construct a 90% confidence interval of a random sample size of 19 cows with a mean of 155.526 and a standard deviation of 19.993. 

• (146.54, 164.52)

• (147.57, 163.48)

• (147.59, 163.46)

• (143.69, 167.36)

The lifetime of a certain type of battery 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)

If you want to construct a confidence interval with a smaller margin of error without changing the sample size, what could you do?

• Increase the confidence level

• Decrease the confidence level

• Increase the variability in the sample

• Decrease the variability in the sample

Percentiles of the bootstrap distribution are provided. Use the percentiles to report a 95% confidence interval for the parameter.

• 6.593 hours to 7.78 hours

• 6.322 hours to 8.082 hours

• 6.438 hours to 7.947 hours

• 6.174 hours to 8.304 hours

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)