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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 proportion who are in agreement with the statement is (0.0841, 0.1151), indicating that there is a 90% probability that the actual proportion of individuals who agree with the statement lies within this range.

Step by step solution

01

Calculate Sample Proportion

First, we calculate the sample proportion (\(p^\)), which is simply the number of individuals who agreed with the statement divided by the total number of individuals in the sample. In this case, \( p^ = \frac{101}{1014} = 0.0996\).
02

Determine the Standard Error

The standard error (SE) for proportion is calculated as \( SE = \sqrt{\frac{{p^*(1-p^)}}{n}}\), where \(p^\) is the sample proportion and \(n\) is the sample size. Substituting into this formula, we get \( SE = \sqrt{\frac{{0.0996* (1-0.0996)}}{1014}} = 0.0094\).
03

Step 3:Identify the z-score for 90% Confidence Interval

For a 90% confidence interval, the z-score (which denotes the number of standard deviations a data-point is from the mean) is approximately 1.645. This value can be found in a standard z-table or it is a generally accepted approximation.
04

Step 4:Calculate the Confidence Interval

Finally, we can calculate the 90% confidence interval using the formula \( CI = p^ 卤 (Z*SE)\), substituting \(p^ = 0.0996\), \( Z = 1.645\), and \( SE = 0.0094\), the confidence interval becomes (0.0996 - (1.645*0.0094), 0.0996 + (1.645*0.0094)) or (0.0841, 0.1151).
05

Interpretation of the Interval

The confidence interval (0.0841, 0.1151) can be interpreted as: We are 90% confident that the true proportion of individuals who agree that landlords should have the right to evict a tenant with AIDS is between 8.41% and 11.51% within the population.

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Key Concepts

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

Sample Proportion
Let's begin by understanding what a sample proportion is and why it matters. In statistics, the sample proportion is the fraction of samples that exhibit a particular trait or characteristic. In our exercise, the sample proportion helps us determine how many individuals agree with the statement compared to the total sample size. For our exercise, the sample proportion is calculated by dividing the number of individuals who agree with the statement by the total number of surveyed individuals.

Here's the formula:
  • Sample Proportion (\(p^\)) = \(\frac{101}{1014}\)
  • Result: \(p^ = 0.0996\)
This means that approximately 9.96% of individuals in the sample agree with the statement.

Sample proportion is crucial because it serves as the foundation for estimating the confidence interval, giving us insights on the population level.
Standard Error
The standard error plays a critical role in the context of confidence intervals. It gives us an idea of how much the sample proportion is expected to vary from the true population proportion. In general, when we calculate the standard error (SE) of a sample proportion, we are assessing the variability of that sample proportion.

Here's the formula for standard error of a proportion:
  • SE = \( \sqrt{\frac{ {p^} (1-p^)}{n}} \)
  • In our case: \( SE = \sqrt{\frac{0.0996*(1-0.0996)}{1014}} \)
  • Result: \( SE = 0.0094 \)
A smaller standard error suggests that the sample proportion is a more accurate reflection of the population proportion.

The standard error acts as a building block for creating a confidence interval by helping define the possible scope of population values.
Z-score
Z-score helps in interpreting the variability in our data by placing it on a standardized scale. In essence, the z-score determines how many standard deviations a sample proportion is away from the mean of the distribution.

For confidence intervals, it plays a crucial role in quantifying the certainty level of our estimation. We can use the standard z-table, or known values for common confidence levels, to find an appropriate z-score.

For a 90% confidence interval, we use:
  • Z-score = 1.645
This number indicates that 90% of the data falls within this many standard deviations of the mean.

Using the z-score in the confidence interval formula helps us incorporate this level of certainty, assuring that our interval truly represents the given confidence level.
Statistical Interpretation
Statistical interpretation of confidence intervals provides a deeper understanding of the analysis and its accuracy. A confidence interval gives a range within which we estimate the true population parameter will fall, given a specific confidence level.

From our previous steps:
  • Calculated Confidence Interval = (0.0841, 0.1151)
This interval suggests that, statistically, we are 90% confident that the true proportion of individuals who agree with the statement lies between 8.41% and 11.51%.

Understanding this output gives us confidence that our sample proportion accurately mirrors the population. This is essential for making informed decisions or conclusions based on sample data.

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Most popular questions from this chapter

searchers in the field of nutrition. The article "Effects of Fast-Food Consumption on Energy Intake and Diet Quality Among Children" (Pediatrics [2004]: \(112-118\) ) reported that 1720 of those in a random sample of 6212 U.S. children indicated that on a typical day they ate fast food. Estimate \(\pi\), the proportion of children in the U.S. who eat fast food on a typical day.

In a survey of 1000 randomly selected adults in the United States, participants were asked what their most favorite and what their least favorite subject was when they were in school (Associated Press, August 17,2005 ). In what might seem like a contradiction, math was chosen more often than any other subject in both categories! Math was chosen by 230 of the 1000 as the favorite subject, and it was also chosen by 370 of the 1000 as the least favorite subject. a. Construct a \(95 \%\) confidence interval for the proportion of U.S. adults for whom math was the favorite subject in school. b. Construct a \(95 \%\) confidence interval for the proportion of U.S. adults for whom math was the least favorite subject.

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.)

"Heinz Plays Catch-up After Under-Filling Ketchup Containers" is the headline of an article that appeared on CNN.com (November 30,2000 ). The article stated that Heinz had agreed to put an extra \(1 \%\) of ketchup into each ketchup container sold in California for a 1 -year period. Suppose that you want to make sure that Heinz is in fact fulfilling its end of the agreement. You plan to take a sample of 20 -oz bottles shipped to California, measure the amount of ketchup in each bottle, and then use the resulting data to estimate the mean amount of ketchup in each bottle. A small pilot study showed that the amount of ketchup in 20-oz bottles varied from \(19.9\) to \(20.3\) oz. How many bottles should be included in the sample if you want to estimate the true mean amount of ketchup to within \(0.1\) oz with \(95 \%\) confidence?

In 1991, California imposed a "snack tax" (a sales 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.
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