91Ó°ÊÓ

A May 2011 Harris Interactive poll asked American adult women, "How often do you think women of your age, who have no special risk factors for breast cancer, should have a mammogram to check for breast cancer?" Fifty percent of women age 40 to 49 years and \(56 \%\) of women age 50 years or older said annually (www.harrisinteractive.com/NewsRoom/PressReleases/tabid/446/ctl/ReadCustomDefault/mid/ \(1506 /\) Articleld \(/ 769 /\) Default.aspx \() .\) Suppose that these results were based on samples of 1055 women age 40 to 49 years and 1240 women age 50 years or older. a. Let \(p_{1}\) and \(p_{2}\) be the proportions of all women age 40 to 49 years and age 50 years or older, respectively, who will say that women of their age with no special risk factors for breast cancer should have an annual mammogram. Construct a \(98 \%\) confidence interval for \(p_{1}-p_{2}\). b. Using a \(1 \%\) significance level, can you conclude that \(p_{1}\) is less than \(p_{2}\) ? Use both the criticalvalue and the \(p\) -value approaches.

Step by step solution

01

Calculate sample proportions

To calculate the sample proportions and their difference, use the given percentage responses and the number of respondents in each group. For the first age group (40 to 49 years old), the sample proportion \( \hat{p_1} \) is \(50% = 0.5\). For the second age group (50 years or older), the sample proportion \( \hat{p_2} \) is \(56% = 0.56\). The difference \( \hat{p_1} - \hat{p_2} = 0.5 - 0.56 = -0.06.\)

02

Construct the Confidence Interval

The 98% confidence interval for \( p_{1}-p_{2} \) is given by \( \hat{p_1} - \hat{p_2} \pm Z_{\alpha/2} \sqrt{\frac{\hat{p_1}(1-\hat{p_1})}{n_1} + \frac{\hat{p_2}(1-\hat{p_2})}{n_2} } \), where \( Z_{\alpha/2} \) is the z-value that captures 98% of the data in the middle of a standard normal distribution (which is approximately 2.33). Substituting the values gives the 98% confidence interval.

03

Carry out the Hypothesis Test

The null hypothesis is \( H_0: p_{1} - p_{2} \geq 0 \) and the alternative hypothesis is \( H_A: p_{1} - p_{2} <0 \). Calculate the test statistics and the p-value. Using the 1% significance level, make a decision regarding the null hypothesis based on the comparison of the p-value and the significance level.

04

Evaluate the Hypothesis using the Critical Value Approach

With the critical value approach, since the test is a left-tailed test, the critical value \( Z_{\alpha} \) corresponding to 1% is -2.33. We reject the null hypothesis if the test statistic is less than -2.33. Based on the calculated test statistics, make a conclusion about the null hypothesis.

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Key Concepts

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

Confidence Interval

A confidence interval is a range of values that estimates an unknown population parameter. In this exercise, we're interested in creating a 98% confidence interval for the difference in proportions, \( p_{1} - p_{2} \), between two groups of women.

The confidence interval is constructed using sample proportions, \( \hat{p_1} = 0.5 \) for women aged 40 to 49, and \( \hat{p_2} = 0.56 \) for women 50 and older.

To calculate, use the formula:

• First, find the difference in sample proportions: \( \hat{p_1} - \hat{p_2} = -0.06 \).
• Determine the standard error (SE) using:
  \[ SE = \sqrt{\frac{\hat{p_1}(1-\hat{p_1})}{n_1} + \frac{\hat{p_2}(1-\hat{p_2})}{n_2} } \] \( n_1 = 1055 \) and \( n_2 = 1240 \).
• Then, calculate the margin of error (ME) by multiplying the SE by the z-value for 98%, about 2.33.

Finally, structure it as:
\[ \hat{p_1} - \hat{p_2} \pm ME \]
This gives you the interval within which the true difference in proportions is expected.

Hypothesis Testing

Hypothesis testing is used to make decisions about population parameters. Here, we're testing if the proportion of younger women \( p_{1} \) saying annual mammograms are needed, is less than that of older women \( p_{2} \).

The hypotheses are:

• Null Hypothesis \( H_0: p_{1} - p_{2} \geq 0 \), implying no difference or \( p_{1} \) is not less.
• Alternative Hypothesis \( H_A: p_{1} - p_{2} < 0 \), implying \( p_{1} \) is less than \( p_{2} \).

Calculate the test statistic using the difference in sample proportions, their standard error, and compare it against the critical z-value or use the p-value. This helps determine if there's enough evidence to reject \( H_{0} \).

In this scenario, if the p-value is less than the significance level, you reject \( H_0 \), concluding \( p_{1} \) is indeed less than \( p_{2} \).

Sample Proportion

The sample proportion is a fundamental concept in statistics, representing the fraction of samples that exhibits a certain characteristic. In this exercise, \( \hat{p_1} \) and \( \hat{p_2} \) are the sample proportions for women aged 40-49 and 50+, respectively, who believe an annual mammogram is necessary.

For each group, compute the sample proportion by dividing the number of "yes" responses by the total sample size.

• For women aged 40-49: \( \hat{p_1} = 0.5 \)
• For women aged 50+: \( \hat{p_2} = 0.56 \)

If a larger sample were used, these sample proportions would likely provide a better estimate of the true population proportions.

Using sample proportions is crucial as they serve as the starting point for calculating test statistics, confidence intervals, and evaluating hypothesis tests.

Significance Level

The significance level, often denoted as \( \alpha \), is a threshold for determining whether a hypothesis test result is statistically significant. In our exercise, we use a 1% significance level, meaning we accept a 1% chance of incorrectly rejecting the null hypothesis (Type I error).

This low level signifies a stringent test, demanding strong evidence to reject \( H_0 \). When comparing the p-value to \( \alpha \),

• If p-value < \( \alpha \), reject \( H_0 \).
• If p-value \geq \( \alpha \), do not reject \( H_0 \).

This decision-making process is critical in statistical inference, ensuring conclusions are made based on solid evidence. A chosen \( \alpha = 0.01 \) safeguards against false positives, which is especially valuable in fields requiring high certainty.