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Chapter 8: Problem 7

Briefly explain the difference between a confidence level and a confidence interval.

Short Answer

Expert verified
A confidence level is the statistical probability that the true population parameter will fall within a certain range of values (a confidence interval). The confidence interval is that range of values itself. Simply put, the level defines the certainty, while the interval defines the range.

Step by step solution

01

Defining Confidence Level

Confidence level refers to the level of certainty we can have that our sample accurately reflects the overall population. It's commonly set at 0.95 or 95%, meaning there is a 95% probability that the range of values within our confidence interval includes the true population parameter.
02

Defining Confidence Interval

On the other hand, confidence interval refers to the range of values within which we can be a specific percent (the confidence level) certain our true population parameter lies. This interval is calculated from the data collected from the sample.
03

Distinguishing Between Confidence Level and Interval

The key difference lies in that confidence level pertains to the degree of certainty about our sample's accurateness, whereas confidence interval deals with the specific range wherein the true population value lies. The confidence interval changes with the sample data, while the confidence level 鈥� often set at 95% 鈥� remains constant.

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Most popular questions from this chapter

An economist wants to find a \(90 \%\) confidence interval for the mean sale price of houses in a state. How large a sample should she select so that the estimate is within \(\$ 3500\) of the population mean? Assume that the standard deviation for the sale prices of all houses in this state is \(\$ 31,500\)

Jack's Auto Insurance Company customers sometimes have to wait a long time to speak to a customer service representative when they call regarding disputed claims. A random sample of 25 such calls yielded a mean waiting time of 22 minutes with a standard deviation of 6 minutes. Construct a \(99 \%\) confidence interval for the population mean of such waiting times. Assume that such waiting times for the population follow a normal distribution.

The U.S. Senate just passed a bill by a vote of \(55-45\) (with all 100 senators voting). A student who took an elementary statistics course last semester says, "We can use these data to make a confidence interval about \(p\). We have \(n=100\) and \(\hat{p}=55 / 100=.55\) " Hence, according to him, a \(95 \%\) confidence interval for \(p\) is $$ \hat{p} \pm z \sigma_{\hat{p}}=.55 \pm 1.96 \sqrt{\frac{(.55)(.45)}{100}}=.55 \pm .098=.452 \text { to } .648 $$ Does this make sense? If not, what is wrong with the student's reasoning?

a. Find the value of \(t\) for the \(t\) distribution with a sample size of 21 and area in the left tail equal to \(.10\). b. Find the value of \(t\) for the \(t\) distribution with a sample size of 14 and area in the right tail equal to \(.025\). c. Find the value of \(t\) for the \(t\) distribution with 45 degrees of freedom and \(.001\) area in the right tail. d. Find the value of \(t\) for the \(t\) distribution with 37 degrees of freedom and \(.005\) area in the left tail.

a. How large a sample should be selected so that the margin of error of estimate for a \(99 \%\) confidence interval for \(p\) is \(.035\) when the value of the sample proportion obtained from a preliminary sample is \(.29\) ? b. Find the most conservative sample size that will produce the margin of error for a \(99 \%\) confidence interval for \(p\) equal to \(.035\).
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