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Chapter 8: Problem 52

A random sample of 16 airline passengers at the Bay City airport showed that the mean time spent waiting in line to check in at the ticket counters was 31 minutes with a standard deviation of 7 minutes. Construct a \(99 \%\) confidence interval for the mean time spent waiting in line by all passengers at this airport. Assume that such waiting times for all passengers are normally distributed.

Short Answer

Expert verified
The 99% confidence interval for the mean wait time is between 26.51 minutes and 35.49 minutes.

Step by step solution

01

Understanding the confidence interval formula

The formula for a confidence interval in this case can be expressed as: \[ \bar{x} \pm Z_{\frac{\alpha}{2}} \times \frac{\sigma}{\sqrt{n}} \] where \(\bar{x}\) is the sample mean, \(Z_{\frac{\alpha}{2}}\) is the critical z-score, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.
02

Look up the critical z-score for a 99% confidence level

Looking at a standard Z-table or using a Z-calculator, the critical Z-value for a 99% confidence interval (0.5% in each tail of the normal distribution) is approximately 2.576.
03

Substitution the known values into the formula

Now, substitute the sample mean (\(\bar{x} = 31\)), sample size (\(n = 16\)), standard deviation (\(\sigma = 7\)), and calculated critical value (\(Z_{\frac{\alpha}{2}} = 2.576\)), into the confidence interval formula: \[ 31 \pm 2.576 \times \frac{7}{\sqrt{16}} \]
04

Compute the confidence interval

Performing the multiplication and addition/subtraction we get our 99% confidence interval: \[ 31 \pm 4.49 \] This implies that we can be 99% confident that the population mean wait time is between 26.51 minutes and 35.49 minutes.

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Most popular questions from this chapter

According to an estimate, \(11 \%\) of junk mail is thrown out without being opened (Source: www. uoregon.edu/ recycle/events_topics_junkmail_text.htm). Suppose that this percentage is based on a random sample of 1400 pieces of junk mail. a. What is the point estimate of the corresponding population proportion? b. Construct a \(95 \%\) confidence interval for the proportion of all pieces of junk mail that is thrown out without being opened. What is the margin of error for this estimate?

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Foods that have less than \(.50\) gram of trans fat per serving are allowed to list the trans fat content as zero grams on labels. A random sample of 32 foods that list the amount of trans fat as zero grams but contain partially hydrogenated oils (the primary source of added trans fat) were evaluated for the amount of trans fat per serving. The sample mean and standard deviation of trans fat were found to be \(.34\) and \(.062\) gram, respectively. Calculate a \(95 \%\) confidence interval for the average amount of trans fat per serving in all foods that are listed as having zero grams of trans fat per serving but contain partially hydrogenated oils.

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Suppose, for a sample selected from a population, \(\bar{x}=25.5\) and \(s=4.9\). a. Construct a \(95 \%\) confidence interval for \(\mu\) assuming \(n=47\). b. Construct a \(99 \%\) confidence interval for \(\mu\) assuming \(n=47\). Is the width of the \(99 \%\) confidence interval larger than the width of the \(95 \%\) confidence interval calculated in part a? If yes, explain why. c. Find a \(95 \%\) confidence interval for \(\mu\) assuming \(n=32\). Is the width of the \(95 \%\) confidence interval for \(\mu\) with \(n=32\) larger than the width of the \(95 \%\) confidence interval for \(\mu\) with \(n=47\) calculated in part a? If so, why? Explain.
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