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Chapter 9: Problem 64

The Gallup Organization conducted a telephone survey on attitudes toward AIDS (Gallup Monthly, 1991). A total of 1014 individuals were contacted. Each individual was asked whether they agreed with the following statement: "Landlords should have the right to evict a tenant from an apartment because that person has AIDS." One hundred one individuals in the sample agreed with this statement. Use these data to construct a \(90 \%\) confidence interval for the proportion who are in agreement with this statement. Give an interpretation of your interval.

Short Answer

Expert verified
The 90% confidence interval for the population proportion who agree with the statement is approximately (0.083, 0.116). This means that with 90% certainty, the proportion of individuals who agree with the statement in the population lies between 8.3% and 11.6%.

Step by step solution

01

Calculation of sample proportion

The goal in the first step is to calculate the sample proportion (\(p\)). The formula to calculate the sample proportion \(p\) is given as \(p = \frac{x}{n}\), where \(x\) is the number of successes and \(n\) is the total number of trials. In this scenario, the number agreeing with the statement (successes) is 101 and the total number of individuals surveyed (trials) is 1014. Substituting these values into the formula yields the approximate proportion of 0.0996.
02

Calculation of standard error

The standard error (\(SE\)) formula for the proportions-based case is \(SE = \sqrt{\frac{p(1 - p)}{n}}\). Substituting the previously calculated value for \(p\) and the known value for \(n\), the standard error for this sample is approximately 0.0097.
03

Calculation of z-score

For a 90% confidence interval, the z-score (which represents how many standard deviations away from the mean the value is) is found in a standard normal distribution table. The z-value is 1.645.
04

Calculation of margin of error

After having the z-score and standard error, the margin of error can be calculated using formula \(ME = z \times SE\). Substituting known values, the calculated margin of error is approximately 0.016.
05

Calculation of confidence interval

Now to form the 90% confidence interval, subtract the margin of error from the sample proportion for the lower limit and add the margin of error to the sample proportion for the upper limit. This interval ranges approximately from 0.083 to 0.116.
06

Interpretation of interval

The final step is to interpret the interval. It can be concluded that there's a 90% confidence that the true population proportion of individuals who agree with the statement lies between 8.3% and 11.6%.

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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 formula used to compute a large-sample confidence interval for \(\pi\) is $$ p \pm(z \text { critical value }) \sqrt{\frac{p(1-p)}{n}} $$ What is the appropriate \(z\) critical value for each of the following confidence levels? a. \(95 \%\) d. \(80 \%\) b. \(90 \%\) e. \(85 \%\) c. \(99 \%\)

Suppose that a random sample of 50 bottles of a par. ticular brand of cough medicine is selected and the alcohol content of each bottle is determined. Let \(\mu\) denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the sample of 50 results in a \(95 \%\) confidence interval for \(\mu\) of \((7.8,9.4)\). a. Would a \(90 \%\) confidence interval have been narrower or wider than the given interval? Explain your answer. b. Consider the following statement: There is a \(95 \%\) chance that \(\mu\) is between \(7.8\) and \(9.4\). Is this statement correct? Why or why not? c. Consider the following statement: If the process of selecting a sample of size 50 and then computing the corresponding \(95 \%\) confidence interval is repeated 100 times, 95 of the resulting intervals will include \(\mu\). Is this statement correct? Why or why not?

In 1991, California imposed a "snack tax" (a sales \(\operatorname{tax}\) on snack food) in an attempt to help balance the state budget. A proposed alternative tax was a \(12 \phi\) -per-pack increase in the cigarette tax. In a poll of 602 randomly selected California registered voters, 445 responded that they would have preferred the cigarette tax increase to the snack tax (Reno Gazette-Journal, August 26,1991 ). Estimate the true proportion of California registered voters who preferred the cigarette tax increase; use a \(95 \%\) confidence interval.

The Center for Urban Transportation Research released a report stating that the average commuting distance in the United States is \(10.9 \mathrm{mi}\) (USA Today, August 13 , 1991). Suppose that this average is actually the mean of a random sample of 300 commuters and that the sample standard deviation is \(6.2 \mathrm{mi}\). Estimate the true mean commuting distance using a \(99 \%\) confidence interval.
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