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Chapter 8: Problem 6

Briefly explain how the width of a confidence interval decreases with a decrease in the confidence level. Give an example.

Short Answer

Expert verified
As confidence level decreases, the corresponding confidence interval narrows. This is because a lower confidence level suggests that there's less need to account for uncertainty. For example, a 95% confidence interval for a population mean might be \(120 卤 6\), but if we lower the confidence to 90%, our confidence interval might narrow to \(120 卤 5\).

Step by step solution

01

Understand Confidence Intervals and Levels

A confidence interval provides an estimated range of values between which an unknown population parameter is likely to lie. The confidence level represents the frequency (in percentage) that the confidence interval contains the true parameter in repeated samples.
02

Understand the Relationship between Confidence Level and Interval Width

A higher confidence level indicates a larger interval as it accommodates more uncertainty about where the population parameter lies. Lowering the confidence level reduces the interval width because there is less need to account for uncertainty. Essentially, less confidence is being expressed that the interval contains the population parameter.
03

Provide an Example

Take a random sample of size \(n= 30\) from a population with standard deviation \(蟽 = 15\). A 95% confidence interval for the population mean might be \(120 卤 6\). If we decrease the confidence level to 90%, our confidence interval might narrow to \(120 卤 5\). In both instances the same mean is used, but the range of values that we are 90% confident the actual value lies within is smaller.

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Most popular questions from this chapter

A Centers for Disease Control and Prevention survey about cell phone use noted that \(14.7 \%\) of U.S. households are wireless-only, which means that the household members use only cell phones and do not have a landline. Suppose that this percentage is based on a random sample of 855 U.S. households (Source: http://www.cdc.gov/nchs/data/nhsr/nhsr014.pdf). a. Construct a \(95 \%\) confidence interval for the proportion of all U.S. households that are wireless-only. b. Explain why we need to construct a confidence interval. Why can we not simply say that \(14.7 \%\) of all U.S. households are wireless-only?

A company randomly selected nine office employees and secretly monitored their computers for one month. The times (in hours) spent by these employees using their computers for non-job-related activities (playing games, personal communications, etc.) during this month are given below. \(\begin{array}{lllllllll}7 & 1 & 29 & 8 & 1 & 14 & 1 & 41 & 6\end{array}\) Assuming that such times for all employees are normally distributed, make a \(95 \%\) confidence interval for the corresponding population mean for all employees of this company.

What is the point estimator of the population proportion, \(p\) ?

A random sample of 36 mid-sized cars tested for fuel consumption gave a mean of \(26.4\) miles per gallon with a standard deviation of \(2.3\) miles per gallon. a. Find a \(99 \%\) confidence interval for the population mean, \(\mu\). b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Describe all possible alternatives. Which alternative is the best and why?

When one is attempting to determine the required sample size for estimating a population mean, and the information on the population standard deviation is not available, it may be feasible to take a small preliminary sample and use the sample standard deviation to estimate the required sample size, \(n .\) Suppose that we want to estimate \(\mu\), the mean commuting distance for students at a community college, to within 1 mile with a confidence level of \(95 \%\). A random sample of 20 students yields a standard deviation of \(4.1\) miles. Use this value of the sample standard deviation, \(s\), to estimate the required sample size, \(n\). Assume that the corresponding population has a normal distribution.
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