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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 article "Consumers Show Increased Liking for Diesel Autos" (USA Today, January 29,2003 ) reported that \(27 \%\) of U.S. consumers would opt for a diesel car if it ran as cleanly and performed as well as a car with a gas engine. Suppose that you suspect that the proportion might be different in your area and that you want to conduct a survey to estimate this proportion for the adult residents of your city. What is the required sample size if you want to estimate this proportion to within \(.05\) with \(95 \%\) confidence? Compute the required sample size first using 27 as a preliminary estimate of \(\pi\) and then using the conservative value of \(.5 .\) How do the two sample sizes compare? What sample size would you recommend for this study?

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?

A study reported in Newsweek (December 23, 1991) involved a sample of 935 smokers. Each individual received a nicotine patch, which delivers nicotine to the bloodstream but at a much slower rate than cigarettes do. Dosage was decreased to 0 over a 12-week period. Suppose that 245 of the subjects were still not smoking 6 months after treatment (this figure is consistent with information given in the article). Estimate the percentage of all smokers who, when given this treatment, would refrain from smoking for at least 6 months.

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 \(n=400\)

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 \(-1.81\)

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