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Chapter 9: Problem 70

The Associated Press (December 16,1991 ) reported that in a random sample of 507 people, only 142 correctly described the Bill of Rights as the first 10 amendments to the U.S. Constitution. Calculate a \(95 \%\) confidence interval for the proportion of the entire population that could give a correct description.

Short Answer

Expert verified
The 95% confidence interval for the proportion of the entire population that could correctly describe the Bill of Rights is calculated to be \(p \pm \)\(Z*\)\(SE\). Fill in the calculated \(p\) and calculated \(SE\) to get the interval.

Step by step solution

01

Calculate the Sample Proportion

The sample proportion \(p\) is calculated as the number of successes (correct responses) divided by the sample size (total responses). Here, the successes are 142 (correct descriptions of the Bill of Rights), and the sample size is 507 people. So, \(p = \frac{142}{507}\).
02

Calculate the Standard Error

The standard error for the sample proportion \(p\) is calculated using the formula \(\sqrt{ \frac{p(1-p)}{n}} \), where \(n\) is the sample size. In this case, \(n = 507\) and \(p\) is the earlier calculated sample proportion.
03

Calculate the Confidence Interval

The formula for the confidence interval is \(p \pm \)\(Z*\)\(SE\), where \(Z*\) is the Z-value from a standard Normal distribution for the desired confidence level (for a 95% confidence interval, \(Z* = 1.96\)), and \(SE\) is the standard error calculated in Step 2. Plug the calculated values into this formula to obtain the confidence interval.

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Most popular questions from this chapter

The authors of the paper "Short-Term Health and Economic Benefits of Smoking Cessation: Low Birth Weight" (Pediatrics [1999]: \(1312-1320\) ) investigated the medical cost associated with babies born to mothers who smoke. The paper included estimates of mean medical cost for low-birth-weight babies for different ethnic groups. For a sample of 654 Hispanic low-birth- weight babies, the mean medical cost was \(\$ 55,007\) and the standard error \((s / \sqrt{n})\) was \(\$ 3011 .\) For a sample of 13 Native American low-birth- weight babies, the mean and standard error were \(\$ 73,418\) and \(\$ 29,577\), respectively. Explain why the two standard errors are so different.

The following data are the calories per half-cup serving for 16 popular chocolate ice cream brands reviewed by Consumer Reports (July 1999): \(\begin{array}{llllllll}270 & 150 & 170 & 140 & 160 & 160 & 160 & 290 \\ 190 & 190 & 160 & 170 & 150 & 110 & 180 & 170\end{array}\) Is it reasonable to use the \(t\) confidence interval to compute a confidence interval for \(\mu\), the true mean calories per half-cup serving of chocolate ice cream? Explain why on why not.

Given a variable that has a \(t\) distribution with the specified degrees of freedom, what percentage of the time will its value fall in the indicated region? a. \(10 \mathrm{df}\), between \(-1.81\) and \(1.81\) b. \(10 \mathrm{df}\), between \(-2.23\) and \(2.23\) c. \(24 \mathrm{df}\), between \(-2.06\) and \(2.06\) d. \(24 \mathrm{df}\), between \(-2.80\) and \(2.80\) e. 24 df, outside the interval from \(-2.80\) to \(2.80\) f. \(24 \mathrm{df}\), to the right of \(2.80\) g. \(10 \mathrm{df}\), to the left of \(-1.81\)

For each of the following choices, explain which would result in a wider large-sample confidence interval for \(\pi\) : a. \(90 \%\) confidence level or \(95 \%\) confidence level b. \(n=100\) or \(n=400\)

The interval from \(-2.33\) to \(1.75\) captures an area of \(.95\) under the \(z\) curve. This implies that another largesample \(95 \%\) confidence interval for \(\mu\) has lower limit \(\bar{x}-2.33 \frac{\sigma}{\sqrt{n}}\) and upper limit \(\bar{x}+1.75 \frac{\sigma}{\sqrt{n}} .\) Would you recommend using this \(95 \%\) interval over the \(95 \%\) interval \(\bar{x} \pm 1.96 \frac{\sigma}{\sqrt{n}}\) discussed in the text? Explain. (Hint: Look at the width of each interval.)
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