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Chapter 6: Problem 272

A \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the paired data in the following table: $$ \begin{array}{lccccc} \hline \text { Case } & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} \\ \hline \text { Treatment 1 } & 22 & 28 & 31 & 25 & 28 \\ \text { Treatment 2 } & 18 & 30 & 25 & 21 & 21 \\ \hline \end{array} $$

Short Answer

Expert verified
The 99% confidence interval for \(\mu_{1}-\mu_{2}\) using the paired data provided is (-1.512, 9.112).

Step by step solution

01

Determine differences for each pair

Subtract the values of Treatment 2 from Treatment 1 for each case to find the differences: \( d = (22-18, 28-30, 31-25, 25-21, 28-21) = (4, -2, 6, 4, 7) \).
02

Calculate the average and standard deviation of differences

Find the average (\(\mu_{d}\)) and standard deviation (\(s_{d}\)) of these differences. Using the formula for average, \(\mu_{d} = \sum d / n\), gives \(\mu_{d} = 19/5 = 3.8\). For the standard deviation, we can use the formula for standard deviation which is \( s_{d} = \sqrt{\frac{\sum (d_{i} - \mu_{d})^{2}}{n-1}} \). Doing this calculation gets us \( s_{d} = \sqrt{\frac{(4-3.8)^2+(-2-3.8)^2+(6-3.8)^2+(4-3.8)^2+(7-3.8)^2}{5-1}} = 3.2619 \).
03

Determine the 99% confidence interval using Z score

The z value for 99% confidence interval is approximately 2.57. So, we can use that to calculate our interval. The confidence interval is given as \(( \mu_{d} - Z \times \frac{s_{d}}{\sqrt{n}}, \mu_{d} + Z \times \frac{s_{d}}{\sqrt{n}} )\), substituting our found values gives us: \(( 3.8 - 2.57 \times \frac{3.2619}{\sqrt{5}}, 3.8 + 2.57 \times \frac{3.2619}{\sqrt{5}} )\), which yields: (-1.512, 9.112)\).

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Key Concepts

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

Paired Sample t-test
The paired sample t-test is a statistical method used to compare the means of two related groups. The classic application is when the same subjects are subjected to two different treatments, like in the exercise. It revolves around the concept that instead of comparing two independent groups, we鈥檙e looking at the difference in outcomes of the same subjects, which can lead to more precise results since it takes into account the variability within the subjects themselves.

For instance, a doctor might want to know whether a new medication is more effective than an old one. By comparing the same group's response to both treatments, the paired sample t-test can determine if there's a statistically significant difference in the effects. In this exercise, we calculated a 99% confidence interval for the mean difference between two treatments, which is a key step in the paired sample t-test. This interval helps to understand the range in which the true mean difference lies with a certain level of certainty.
Standard Deviation
Standard deviation (SD) is a measurement that tells us how spread out the numbers are in a set of data. It is the square root of the variance, which is the average of the squared differences from the Mean. A low SD indicates that the data points tend to be close to the mean, while a high SD indicates that the data points are spread out over a wider range of values.

Considering our exercise, the SD of the differences between treatments represents how much variability there is in the response to the treatments. The calculation of the SD is crucial as it plays a direct role in constructing confidence intervals. In the context of a paired sample t-test, it helps to assess the consistency of the treatment effects. For educators, it's key to emphasize that while understanding the formula is important, grasping the concept of variability that SD represents is equally critical for students.
Difference of Means
The difference of means is a term used to describe the arithmetic difference between the averages of two data sets. When we speak about a paired sample t-test, we are particularly interested in the difference of means from the same subjects under two conditions, as this indicates whether there was a significant change or effect due to the treatment or condition.

In our textbook example, after finding the difference for each data pair and the mean of these differences, we get an average effect. This mean difference, represented by \( \mu_d \) in our steps, is the cornerstone of our hypothesis testing in a paired t-test. It provides a single number that summarizes the comparative effect of the two treatments. Understanding the difference of means in the context of paired comparisons helps students grasp why we can鈥檛 just compare averages directly when the same subjects are involved in both datasets.

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Most popular questions from this chapter

A margin of error within \(\pm 2 \%\) with \(95 \%\) confidence. An initial small sample has \(\hat{p}=0.78\).

Exercise B.5 on page 305 introduces a study examining the effect of diet cola consumption on calcium levels in women. A sample of 16 healthy women aged 18 to 40 were randomly assigned to drink 24 ounces of either diet cola or water. Their urine was collected for three hours after ingestion of the beverage and calcium excretion (in mg) was measured. The summary statistics for diet cola are \(\bar{x}_{C}=56.0\) with \(s_{C}=4.93\) and \(n_{C}=8\) and the summary statistics for water are \(\bar{x}_{W}=49.1\) with \(s_{W}=3.64\) and \(n_{W}=8\). Figure 6.26 shows dotplots of the data values. Test whether there is evidence that diet cola leaches calcium out of the system, which would increase the amount of calcium in the urine for diet cola drinkers. In Exercise B.5, we used a randomization distribution to conduct this test. Use a t-distribution here, after first checking that the conditions are met and explaining your reasoning. The data are stored in ColaCalcium.

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? $$ \begin{array}{rrcl} \hline \text { Year } & \text { Favor } & \text { Oppose } & \text { Total } \\\ \hline 1974 & 937 & 473 & 1410 \\ 2006 & 1945 & 870 & 2815 \\ \hline \end{array} $$

Refer 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 invasive breast cancer? $$ \begin{array}{lcc} \hline \text { Condition } & \text { HRT Group } & \text { Placebo Group } \\ \hline \text { Cardiovascular Disease } & 164 & 122 \\ \text { Invasive Breast Cancer } & 166 & 124 \\ \text { Cancer (all) } & 502 & 458 \\ \text { Fractures } & 650 & 788 \\ \hline \end{array} $$

Use a t-distribution. Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the endpoints of the t-distribution with \(2.5 \%\) beyond them in each tail if the samples have sizes \(n_{1}=15\) and \(n_{2}=25\)
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