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Chapter 6: Problem 176

Has Support for Capital Punishment Changed over Time? The General Social Survey (GSS) has been collecting demographic, behavioral, and attitudinal information since 1972 to monitor changes within the US and to compare the US to other nations. \({ }^{46}\) Support for capital punishment (the death penalty) in the US is shown in 1974 and in 2006 in the two-way table in Table \(6.6 .\) Find a \(95 \%\) confidence interval for the change in the proportion supporting capital punishment between 1974 and 2006. Is it plausible that the proportion supporting capital punishment has not changed? $$ \begin{array}{rrcl} \hline \text { Year } & \text { Favor } & \text { Oppose } & \text { Total } \\\ \hline 1974 & 937 & 473 & 1410 \\ 2006 & 1945 & 870 & 2815 \\ \hline \end{array} $$

Short Answer

Expert verified
The estimated change in the proportion supporting capital punishment from 1974 to 2006 is 0.025, and the 95% confidence interval for this change is -0.019 to 0.069. Since this interval includes zero, it is plausible that the proportion supporting capital punishment has not changed.

Step by step solution

01

Calculate the Proportions

First we will calculate the proportions of people who support capital punishment for each year. The proportion is calculated by dividing the number of people who favor by the total number of people. For 1974, this will be \( \frac{937}{1410} = 0.665 \) and for 2006, \( \frac{1945}{2815} = 0.690 \). The change in proportions is calculated by subtracting the 1974 proportion from 2006 proportion, which results in \( 0.690 - 0.665 = 0.025 \). So, there was a 2.5% increase in support for capital punishment from 1974 to 2006.
02

Calculate the Standard Error

Next, we will calculate the standard error (SE) using the formula for standard error of proportions. The formula is: \[ SE_{\Delta} = \sqrt{\frac{p1(1 - p1)}{n1} + \frac{p2(1 - p2)}{n2}} \] Where: - \( p1 \) and \( p2 \) are the proportions in each sample - \( n1 \) and \( n2 \) are the sizes of each sample Using the given numbers, we calculate as follows: \[ SE_{\Delta} = \sqrt{\frac{0.665(1 - 0.665)}{1410} + \frac{0.690(1 - 0.690)}{2815}} = 0.022 \] The standard error of the difference is 0.0225.
03

Establish the Confidence Interval

Now, we can calculate the 95% confidence interval. The formula for a confidence interval is \( \text{estimate} \pm z * SE_{\text{estimate}} \), where z is the number of standard deviations you want to extend from the mean (1.96 for a 95% confidence interval). Thus, the confidence interval is: \( 0.025 \pm 1.96 * 0.022 \), which means it is between -0.019 and 0.069.
04

Interpret the Confidence Interval

The confidence interval runs from -0.019 to 0.069. Since this interval contains zero, we can say that it is plausible that the proportion supporting capital punishment has not changed, because zero would correspond to no change.

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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 provides a range of values that helps us understand the extent of uncertainty around a sample statistic like a mean or proportion. In our exercise, we are calculating a 95% confidence interval to estimate the change in support for capital punishment between 1974 and 2006. This means that we are 95% confident that the true change in proportion lies within this interval. To set up a confidence interval, we use the equation:
\[ \text{Estimate} \pm z \times SE_{\text{estimate}} \]
In a 95% confidence interval, the value of \(z\) is typically 1.96 since it covers approximately 95% of a standard normal distribution. The confidence interval helps in making informed decisions by considering the range within which the true proportion change might fall, while addressing the variability inherent in sampling.
Proportions
Proportions are a way to express how a part relates to a whole. They are especially useful in surveys and studies to quantify particular characteristics, like support or opposition to an idea. In our case, the proportion of people supporting capital punishment in two different years was calculated. A proportion is mathematically expressed as:
\[ p = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Outcomes}} \]
For instance, the proportion supporting capital punishment in 1974 was calculated as \( \frac{937}{1410} = 0.665 \). Calculating the proportion lets us see the part of the population that supports capital punishment over time. The difference between these proportions allows us to identify trends and changes in societal attitudes.
Standard Error
The standard error (SE) quantifies how much a sample statistic, like a mean or proportion, is expected to vary from the true population parameter. It reflects the degree of sampling variability and is crucial in constructing confidence intervals.
In the context of comparing proportions, as in our example, we calculate the standard error of the difference between two proportions using:
\[ SE_{\Delta} = \sqrt{\frac{p1(1 - p1)}{n1} + \frac{p2(1 - p2)}{n2}} \]
Here, \(p1\) and \(p2\) are the sample proportions, while \(n1\) and \(n2\) represent the sample sizes. Understanding standard error helps us grasp how "off" our sample could be from the actual population values, and it serves as the baseline for constructing confidence intervals, allowing us to make reliable inferences from our sample data.

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Most popular questions from this chapter

Of the 50 states in the Unites States, Alaska has the largest percentage of males and Rhode Island has the largest percentage of females. (Interestingly, Alaska is the largest state and Rhode Island is the smallest). According to the 2010 US Census, the population of Alaska is \(52.0 \%\) male and the population of Rhode Island is \(48.3 \%\) male. If we randomly sample 300 people from Alaska and 300 people from Rhode Island, what is the approximate distribution of \(\hat{p}_{a}-\hat{p}_{r i}\), where \(\hat{p}_{a}\) is the proportion of males in the Alaskan sample and \(\hat{p}_{r i}\) is the proportion of males in the Rhode Island sample?

Rrefer to a study on hormone replacement therapy. Until 2002 , hormone replacement therapy (HRT), taking hormones to replace those the body no longer makes after menopause, was commonly prescribed to post-menopausal women. However, in 2002 the results of a large clinical trial \(^{56}\) were published, causing most doctors to stop prescribing it and most women to stop using it, impacting the health of millions of women around the world. In the experiment, 8506 women were randomized to take HRT and 8102 were randomized to take a placebo. Table 6.16 shows the observed counts for several conditions over the five years of the study. (Note: The planned duration was 8.5 years. If Exercises 6.205 through 6.208 are done correctly, you will notice that several of the p-values are just below \(0.05 .\) The study was terminated as soon as HRT was shown to significantly increase risk (using a significance level of \(\alpha=0.05)\), because at that point it was unethical to continue forcing women to take HRT). Does HRT influence the chance of a woman getting cardiovascular disease? $$ \begin{array}{lcc} \hline \text { Condition } & \text { HRT Group } & \text { Placebo Group } \\ \hline \text { Cardiovascular Disease } & 164 & 122 \\ \text { Invasive Breast Cancer } & 166 & 124 \\ \text { Cancer (all) } & 502 & 458 \\ \text { Fractures } & 650 & 788 \\ \hline \end{array} $$

Random samples of the given sizes are drawn from populations with the given means and standard deviations. For each scenario: (a) Find the mean and standard error of the distribution of differences in sample means, \(\bar{x}_{1}-\bar{x}_{2}\) (b) If the sample sizes are large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 100 from Population 1 with mean 87 and standard deviation 12 and samples of size 80 from Population 2 with mean 81 and standard deviation 15

The dataset BaseballHits gives 2010 season statistics for all Major League Baseball teams. We treat this as a sample of all MLB teams in all years. Computer output of descriptive statistics for the variable giving the batting average is shown: $$ \begin{aligned} &\text { Descriptive Statistics: BattingAvg }\\\ &\begin{array}{lrrrrr} \text { Variable } & \mathrm{N} & \mathrm{N}^{*} & \text { Mean } & \text { SE Mean } & \text { StDev } \\ \text { BattingAvg } & 30 & 0 & 0.25727 & 0.00190 & 0.01039 \\ \text { Minimum } & & \text { Q1 } & \text { Median } & \text { Q3 } & \text { Maximum } \\ 0.23600 & 0.24800 & 0.25700 & 0.26725 & 0.27600 \end{array} \end{aligned} $$ (a) How many teams are included in the dataset? What is the mean batting average? What is the standard deviation? (b) Use the descriptive statistics above to conduct a hypothesis test to determine whether there is evidence that average team batting average is different from \(0.250 .\) Show all details of the test. (c) Compare the test statistic and p-value you found in part (b) to the computer output below for the same data:

Impact of Sample Size on Accuracy Compute the standard error for sample proportions from a population with proportion \(p=0.4\) for sample sizes of \(n=30, n=200,\) and \(n=1000 .\) What effect does increasing the sample size have on the standard error? Using this information about the effect on the standard error, discuss the effect of increasing the sample size on the accuracy of using a sample proportion to estimate a population proportion.
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