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Government Waste In a Gallup poll 513 national adults aged 18 years or older who consider themselves to be Republican were asked, "Of every tax dollar that goes to the federal government in Washington, D.C., how many cents of each dollar would you say are wasted?" The mean wasted was found to be 54 cents with a standard deviation of 2.9 cents. The same question was asked of 513 national adults aged 18 years or older who consider themselves to be Democrat. The mean wasted was found to be 41 cents with a standard deviation of 2.6 cents. Construct a \(95 \%\) confidence interval for the mean difference in government waste, \(\mu_{R}-\mu_{D} .\) Interpret the interval.

Short Answer

Expert verified
The 95% confidence interval for the mean difference in perceived government waste is (12.663, 13.337) cents per dollar.

Step by step solution

01

- Identify Given Data

Identify and list down all the given data from the problem. For Republicans: \(\bar{X}_R = 54\) cents, \(s_R = 2.9\) cents, \(n_R = 513\). For Democrats: \(\bar{X}_D = 41\) cents, \(s_D = 2.6\) cents, \(n_D = 513\).
02

- Determine the Difference of Means

Calculate the difference of sample means: \( \bar{X}_R - \bar{X}_D = 54 - 41 = 13 \) cents
03

- Calculate the Standard Error of the Difference in Means

Use the standard deviations and sample sizes to calculate the standard error:\[ SE = \sqrt{ \frac{s_R^2}{n_R} + \frac{s_D^2}{n_D} } \] \[ SE = \sqrt{ \frac{2.9^2}{513} + \frac{2.6^2}{513} } \] \[ SE = \sqrt{ \frac{8.41}{513} + \frac{6.76}{513} } \] \[ SE = \sqrt{ \frac{8.41 + 6.76}{513} } \] \[ SE = \sqrt{ \frac{15.17}{513} } \] \[ SE = \sqrt{0.0296} \approx 0.172 \]
04

- Determine the Critical Value

For a 95% confidence interval, the critical value (\(z^*\)) for a Z-distribution is 1.96.
05

- Construct the Confidence Interval

Use the formula for the confidence interval. The confidence interval is given by: \( (\bar{X}_R - \bar{X}_D) \pm z^* \cdot SE \) \[ 13 \pm 1.96 \cdot 0.172 \] \[ 13 \pm 0.337 \] This results in the interval: \[ (13 - 0.337, 13 + 0.337) \] \[ (12.663, 13.337) \]
06

- Interpret the Interval

With 95% confidence, it can be stated that the true mean difference in perceived government waste between Republicans and Democrats is between 12.663 cents and 13.337 cents per dollar.

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Key Concepts

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

Mean Difference
The mean difference is a way to compare the average values between two groups. In this exercise, the two groups are Republicans and Democrats. The mean wasted cents for Republicans is 54, while for Democrats it is 41. To find out how much more Republicans think is wasted compared to Democrats, we calculate the mean difference. We simply subtract the Democrat mean from the Republican mean: 54 - 41 = 13 cents. This tells us that on average, Republicans believe 13 more cents per tax dollar are wasted compared to Democrats. Understanding the mean difference is helpful in comparing groups and identifying any significant variations in their responses.
Standard Error
Standard error measures how much the sample mean of a group is expected to vary. In this problem, we need to determine the standard error of the difference in mean wasted cents between Republicans and Democrats. The formula we use is: \[ SE = \sqrt{ \frac{s_R^2}{n_R} + \frac{s_D^2}{n_D} } \]. Here, \(s_R\) and \(s_D\) are the standard deviations for Republicans and Democrats, while \(n_R\) and \(n_D\) are their respective sample sizes.
Calculating step-by-step:
1. Square the standard deviations:
For Republicans: 2.9^2 = 8.41
For Democrats: 2.6^2 = 6.76
2. Divide by the sample sizes:
For Republicans: 8.41/513
For Democrats: 6.76/513
3. Sum these two values and take the square root:
\[ SE = \sqrt{ \frac{8.41}{513} + \frac{6.76}{513} } = \sqrt{0.0296} \approx 0.172 \]
This combined standard error helps us understand how much the mean difference (13 cents) might vary due to sampling variability.
Critical Value
The critical value is a threshold number that helps us determine the extent to which we can be confident in our interval. For the 95% confidence level used in the exercise, the critical value for a Z-distribution is 1.96. This means, statistically, that there is a 95% chance that the true mean difference lies within 1.96 standard errors of the observed mean difference.
To construct our 95% confidence interval, we multiply the critical value by the standard error and add/subtract this product from our mean difference:
\[ \text{Confidence interval} = (\bar{X}_R - \bar{X}_D) \pm z^* \times SE = 13 \pm 1.96 \times 0.172 = 13 \pm 0.337 \]
Thus, our confidence interval is (12.663, 13.337). This range tells us that we can be 95% confident that the true mean difference in perceived government waste between Republicans and Democrats lies within this interval. Interpreting this, we understand that there is a statistically significant difference was believed to be wasted.

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Most popular questions from this chapter

A quality-control engineer wants to find out whether or not a new machine that fills bottles with liquid has less variability than the machine currently in use. The engineer calibrates each machine to fill bottles with 16 ounces of a liquid. After running each machine for 5 hours, she randomly selects 15 filled bottles from each machine and measures their contents. She obtains the following results: $$ \begin{array}{ccc|ccc}& {\text { Old Machine }} & && {\text { New Machine }} \\ \hline 16.01 & 16.04 & 15.96 & 16.02 & 15.96 & 16.05 \\ \hline 16.00 & 16.07 & 15.89 & 15.95 & 15.99 & 16.02 \\ \hline 16.04 & 16.05 & 15.91 & 16.00 & 15.97 & 16.03 \\ \hline 16.10 & 16.01 & 16.00 & 16.06 & 16.05 & 15.94 \\ \hline 15.92 & 16.16 & 15.92 & 16.08 & 15.96 & 15.95 \\ \hline \end{array} $$ (a) Is the variability in the new machine less than that of the old machine at the \(\alpha=0.05\) level of significance? Note: Normal probability plots indicate that the data are normally distributed. (b) Draw boxplots of each data set to confirm the results of part (a) visually.

MythBusters In a MythBusters episode, the question was asked,"Which is better? A four-way stop or a roundabout?" "Better" was determined based on determining the number of vehicles that travel through the four-way stop over a 5 -minute interval of time. Suppose the folks at MythBusters conducted this experiment 15 times for each intersection design. (a) What is the variable of interest in this study? Is it qualitative or quantitative? (b) Explain why the data might be analyzed by comparing two independent means. Include in this explanation what the null and alternative hypotheses would be and what each mean represents (c) A potential improvement on the experimental design might be to identify 15 groups of different drivers and ask each group to drive through each intersection design for the 5 - minute time interval. Explain why this is a better design and explain the role randomization would play. What would be the null and alternative hypotheses here and how would the mean be computed?

On April \(12,1955,\) Dr. Jonas Salk released the results of clinical trials for his vaccine to prevent polio. In these clinical trials, 400,000 children were randomly divided in two groups. The subjects in group 1 (the experimental group) were given the vaccine, while the subjects in group 2 (the control group) were given a placebo. Of the 200,000 children in the experimental group, 33 developed polio. Of the 200,000 children in the control group, 115 developed polio. (a) What type of experimental design is this? (b) What is the response variable? (c) What are the treatments? (d) What is a placebo? (e) Why is such a large number of subjects needed for this study? (f) Does it appear to be the case that the vaccine was effective?

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Construct a confidence interval for \(p_{1}-p_{2}\) at the given level of confidence. \(x_{1}=28, n_{1}=254, x_{2}=36, n_{2}=301,95 \%\) confidence

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