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Chapter 8: Problem 39

A random sample of students who took the SAT college entrance examination twice found that 427 of the respondents had paid for coaching courses and that the remaining 2733 had not. \({ }^{14}\) Construct and interpret a \(99 \%\) confidence interval for the proportion of coaching among students who retake the SAT.

Short Answer

Expert verified
The 99% confidence interval is approximately [0.1207, 0.1495].

Step by step solution

01

Identify sample sizes and proportions

From the given data, we have two groups: students who took coaching courses (427) and those who did not (2733). The total sample size is therefore 427+2733 = 3160. We need to find the proportion of students who took coaching, estimated by dividing the number of coached students by the total number of students, so \( \hat{p} = \frac{427}{3160} \approx 0.1351 \).
02

Determine the critical value

For a 99% confidence interval, we need the critical value \( z \) which corresponds to the significance level of \( \alpha = 0.01 \). Using the Z-table, find \( z_{\alpha/2} \) where \( \alpha/2 = 0.005 \). This value is approximately 2.576.
03

Calculate the standard error

The standard error \( SE \) of the sample proportion is calculated using the formula \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \) where \( n = 3160 \). Substituting the values, we have \( SE = \sqrt{\frac{0.1351 \times (1-0.1351)}{3160}} \approx 0.0056 \).
04

Construct the confidence interval

The confidence interval is found using the formula: \( \hat{p} \pm z \times SE \). Thus, it becomes \( 0.1351 \pm 2.576 \times 0.0056 \approx 0.1351 \pm 0.0144 \). Calculating, we get the interval: \( [0.1207, 0.1495] \).
05

Interpret the confidence interval

The 99% confidence interval for the proportion of students who took coaching is approximately \([0.1207, 0.1495]\). This means we are 99% confident that the true proportion of students who took coaching courses ranges from 12.07% to 14.95%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Error
The Standard Error (SE) is a measure that indicates the accuracy of a sample proportion as an estimate of the true population proportion. It is similar to the concept of standard deviation for a single data point.

The formula to calculate the standard error for a sample proportion is: - \[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]- where \( \hat{p} \) is the sample proportion and \( n \) is the total sample size. - This calculation shows how much the proportion \

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Most popular questions from this chapter

A national opinion poll found that \(44 \%\) of all American adults agree that parents should be given vouchers that are good for education at any public or private school of their choice. The result was based on a small sample. (a) How large an SRS is required to obtain a margin of error of 0.03 (that is, \(\pm 3 \%\) ) in a \(99 \%\) confidence interval? Answer this question using the previous poll's result as the guessed value for \(\hat{p}\). (b) Answer the question in part (a) again, but this time use the conservative guess \(\hat{p}=0.5 .\) By how much do the two sample sizes differ?

Determine the point estimator you would use and calculate the value of the point estimate. How many pairs of shoes, on average, do female teens have? To find out, an AP Statistics class conducted a survey. They selected an SRS of 20 female students from their school. Then they recorded the number of pairs of shoes that each student reported having. Here are the data: $$\begin{array}{rrrrrrrrrr}50 & 26 & 26 & 31 & 57 & 19 & 24 & 22 & 23 & 38 \\\13 & 50 & 13 & 34 & 23 & 30 & 49 & 13 & 15 & 51\end{array}$$

A researcher plans to use a random sample of families to estimate the mean monthly family income for a large population. The researcher is deciding between a \(95 \%\) confidence level and a \(99 \%\) confidence level. Compared to a \(95 \%\) confidence interval, a \(99 \%\) confidence interval will be (a) narrower and would involve a larger risk of being incorrect. (b) wider and would involve a smaller risk of being incorrect. (c) narrower and would involve a smaller risk of being incorrect. (d) wider and would involve a larger risk of being incorrect. (e) wider and would have the same risk of being incorrect.

Have efforts to promote equality for women gone far enough in the United States? A poll on this issue by the cable network MSNBC contacted 1019 adults. A newspaper article about the poll said, "Results have a margin of sampling error of plus or minus 3 percentage points." (a) The news article said that \(65 \%\) of men, but only \(43 \%\) of women, think that efforts to promote equality have gone far enough. Explain why we do not have enough information to give confidence intervals for men and women separately. (b) Would a \(95 \%\) confidence interval for women alone have a margin of error less than \(0.03,\) about equal to \(0.03,\) or greater than 0.03 ? Why? (You see that the news article's statement about the margin of error for poll results is a bit misleading.)

What proportion of students are willing to report cheating by other students? A student project put this question to an SRS of 172 undergraduates at a large university: "You witness two students cheating on a quiz. Do you go to the professor?" Only 19 answered "Yes." \({ }^{12}\) (a) Identify the population and parameter of interest. (b) Check conditions for constructing a confidence interval for the parameter. (c) Construct a \(99 \%\) confidence interval for \(p .\) Show your method. (d) Interpret the interval in context.
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