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Chapter 8: Problem 21

Fear of breast cancer A recent survey of 1000 American women between the ages of 45 and 64 asked them what medical condition they most feared. Of those sampled, \(61 \%\) said breast cancer, \(8 \%\) said heart disease, and the rest picked other conditions. By contrast, currently about \(3 \%\) of female deaths are due to breast cancer, whereas \(32 \%\) are due to heart disease. \(^{5}\) a. Construct a \(90 \%\) confidence interval for the population proportion of women who most feared breast cancer. Interpret. b. Indicate the assumptions you must make for the inference in part a to be valid.

Short Answer

Expert verified
The 90% confidence interval is (58.48%, 63.52%). Assumptions: random sampling and normal approximation applicability.

Step by step solution

01

Understand the Problem

We need to calculate a 90% confidence interval for the population proportion of women who most feared breast cancer based on the sample data provided. Additionally, we need to identify assumptions for this inference.
02

Identify Sample Proportion

From the survey, 61% of 1000 women feared breast cancer. Therefore, the sample proportion \(\hat{p}\) is 0.61.
03

Determine Standard Error

The standard error for a proportion is calculated using the formula \(SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\), where \(n = 1000\). Thus, \(SE = \sqrt{\frac{0.61(1-0.61)}{1000}} = \sqrt{\frac{0.61 \times 0.39}{1000}} \approx 0.0153\).
04

Find the Z-score for 90% Confidence

For a 90% confidence interval, the Z-score corresponding to the middle 90% is approximately 1.645 (from standard normal distribution tables).
05

Calculate Confidence Interval

The confidence interval is given by \(\hat{p} \pm Z \cdot SE\). Substitute the values: \(0.61 \pm 1.645 \times 0.0153\). Thus, the confidence interval is \( (0.61 - 0.0252, 0.61 + 0.0252) = (0.5848, 0.6352) \).
06

Interpret the Confidence Interval

We can say with 90% confidence that the true population proportion of American women aged 45-64 who most fear breast cancer lies between 58.48% and 63.52%.
07

State Assumptions for Validity

To be valid, the following assumptions must hold: (1) The sample of 1000 women is a simple random sample of the population. (2) The sample size is large enough to use the normal approximation, which is typically satisfied if both \(np\) and \(n(1-p)\) are greater than 5.

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Key Concepts

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

Sample Proportion
The concept of a sample proportion is vital in statistics, especially when estimating a characteristic of a large population from a small group. In the exercise, out of 1000 surveyed women, 61% expressed fear of breast cancer, meaning
  • our sample proportion, denoted as \( \hat{p} \), is 0.61.
Sample proportions help us make inferences about the overall population, acting as estimates to see how a certain opinion or characteristic of interest might apply broadly.
To compute the sample proportion, divide the number of individuals with the trait by the total number of individuals in the sample.
This proportion is a key ingredient in building a confidence interval, which helps predict the range where the true population proportion lies.
Standard Error
The standard error gives us a measure of the variability of the sample proportion in relation to the population proportion. It tells us how much the sample proportion \( \hat{p} \) is likely to fluctuate from the true population proportion.
The formula for standard error of a sample proportion is \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \), where \( n \) is the sample size.
  • With a sample size of 1000 and a sample proportion of \( \hat{p} = 0.61 \), the standard error calculates to approximately 0.0153.
A smaller standard error implies that our estimate is more precise. The smaller the standard error, the closer our sample proportion is to the true population proportion and the narrower our confidence interval will be.
Z-score
The Z-score arises when we need to calculate confidence intervals. It relates to how many standard deviations an element is from the mean.
In this context, it's used to identify the margin of error for a chosen confidence level. For a 90% confidence interval, we use a Z-score of approximately 1.645.
  • This Z-score means we expect 90% of our sample proportions to fall within this interval in multiple samples.
To apply it, multiply the Z-score by the standard error to get the range of the confidence interval around the sample proportion. This results in a balanced spread equally below and above \( \hat{p} \).
Assumptions for Validity
Assumptions are crucial to ensure our confidence interval is valid and reliable.
For this exercise, the following assumptions need to be checked:
  • **Simple Random Sample:** The sample must be randomly selected from the population to avoid biases and representation errors.
  • **Sample Size Adequacy:** For the normal approximation to hold, the sample size should match the condition that both \( np \) and \( n(1-p) \) are greater than 5, ensuring a sufficient amount of successes and failures for credibility.
These assumptions reinforce that the formula for the confidence interval is applied correctly, providing a dependable estimation.

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Most popular questions from this chapter

Multiple choice: Number of close friends \(\quad\) Based on responses of 1467 subjects in a General Social Survey, a \(95 \%\) confidence interval for the mean number of close friends equals \((6.8,8.0) .\) Which \(t w o\) of the following interpretations are correct? a. We can be \(95 \%\) confident that \(\bar{x}\) is between 6.8 and 8.0 . b. We can be \(95 \%\) confident that \(\mu\) is between 6.8 and 8.0 . c. Ninety-five percent of the values of \(X=\) number of close friends (for this sample) are between 6.8 and \(8.0 .\) d. If random samples of size 1467 were repeatedly selected, then \(95 \%\) of the time \(\bar{x}\) would be between 6.8 and \(8.0 .\) e. If random samples of size 1467 were repeatedly selected, then in the long run \(95 \%\) of the confidence intervals formed would contain the true value of \(\mu\).

Why bootstrap? Explain the purpose of using the bootstrap method.

Wife doesn't want kids The 1996 GSS asked, "If the husband in a family wants children, but the wife decides that she does not want any children, is it all right for the wife to refuse to have children?" Of 699 respondents, 576 said yes. a. Find a \(99 \%\) confidence interval for the population proportion who would say yes. Can you conclude that the population proportion exceeds \(75 \%\) ? Why? b. Without doing any calculation, explain whether the interval in part a would be wider or narrower than a \(95 \%\) confidence interval for the population proportion who would say yes.

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