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Listed below are body temperatures from five different subjects measured at \(8 \mathrm{AM}\) and again at \(12 \mathrm{AM}\) (from Data Set 3 "Body Temperatures" in Appendix B). Find the values of \(\bar{d}\) and \(s_{d}\). In general, what does \(\mu_{d}\) represent? $$\begin{array}{l|l|l|l|l|l} \hline \text { Temperature ( } \text { ( } \text { F) at 8 AM } & 97.8 & 99.0 & 97.4 & 97.4 & 97.5 \\ \hline \text { Temperature (FF) at 12 AM } & 98.6 & 99.5 & 97.5 & 97.3 & 97.6 \\\ \hline \end{array}$$

Short Answer

Expert verified
\(\bar{d} = 0.28\) and \(s_d \approx 0.36\). \(\mu_d\) is the population mean of the differences in body temperatures.

Step by step solution

01

Calculate the Differences

Find the difference between the body temperatures measured at 8 AM and 12 AM for each subject. Let these differences be denoted as \(d_i\). Use the formula: \(d_i = \text{Temperature at 12 AM} - \text{Temperature at 8 AM}\).
02

List the Differences

Compute the differences for each subject: Subject 1: \(d_1 = 98.6 - 97.8 = 0.8\)Subject 2: \(d_2 = 99.5 - 99.0 = 0.5\)Subject 3: \(d_3 = 97.5 - 97.4 = 0.1\)Subject 4: \(d_4 = 97.3 - 97.4 = -0.1\)Subject 5: \(d_5 = 97.6 - 97.5 = 0.1\)
03

Calculate the Mean Difference \(\bar{d}\)

To find \(\bar{d}\), calculate the average of the differences. Use the formula: \(\bar{d} = \frac{\sum_{i=1}^{n} d_i}{n}\). Substitute the values: \(\bar{d} = \frac{0.8 + 0.5 + 0.1 + (-0.1) + 0.1}{5} = \frac{1.4}{5} = 0.28\)
04

Calculate the Variance

To find the variance of the differences, use the formula: \(s_{d}^{2} = \frac{\sum_{i=1}^{n} (d_i - \bar{d})^2}{n-1}\). First, compute each term \(d_i - \bar{d}\):Subject 1: \(0.8 - 0.28 = 0.52\)Subject 2: \(0.5 - 0.28 = 0.22\)Subject 3: \(0.1 - 0.28 = -0.18\)Subject 4: \(-0.1 - 0.28 = -0.38\)Subject 5: \(0.1 - 0.28 = -0.18\). Square each term and sum them: \((0.52^2 + 0.22^2 + (-0.18)^2 + (-0.38)^2 + (-0.18)^2) = 0.2704 + 0.0484 + 0.0324 + 0.1444 + 0.0324 = 0.528\). Now, divide this by \(n-1 = 4\): \(s_{d}^{2} = \frac{0.528}{4} = 0.132\)
05

Calculate the Standard Deviation \(s_d\)

Take the square root of the variance to find the standard deviation: \(s_d = \sqrt{s_{d}^{2}} = \sqrt{0.132} \approx 0.36\).
06

Interpret \(\mu_d\)

In general, \(\mu_d\) represents the population mean of the differences in body temperatures between 8 AM and 12 AM for the entire population from which the sample is derived.

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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, denoted as \(\bar{d}\), is a simple yet essential concept in statistical analysis. When comparing two sets of measurements taken from the same subjects, like body temperatures at two different times, we calculate the difference for each subject first. For this problem, differences in body temperatures measured at 8 AM and 12 AM are computed as shown in Step 1 and 2. All these differences are then averaged to find the mean difference. The formula for the mean difference is: \[\bar{d} = \frac{\sum_{i=1}^{n} d_i}{n}\],\ where \(d_i\) denotes the difference for each subject, and \(n\) is the total number of subjects. In this exercise,\ \(\bar{d}\) was found to be 0.28. This average provides insight into how much temperatures typically change from morning to noon.
standard deviation
Standard deviation, denoted as \(s_d\), takes the analysis a step further by quantifying how spread out the differences are around the mean difference. After finding the mean difference, the formula for calculating the standard deviation \(s_d\) is: \[s_d = \sqrt{s_{d}^{2}}\].\ This involves first calculating the variance \[s_{d}^{2}\],\ then taking its square root. It tells us if the differences in temperatures are consistent across subjects or if there's a high degree of variability. In our problem, we found \(s_d\) to be approximately 0.36. A smaller standard deviation would imply that most differences are close to 0.28, while a larger standard deviation indicates more variability.
sample variance
Sample variance \(s_{d}^{2}\) is a measure of the dispersion of the differences within our sample data. It's calculated before finding the standard deviation. The formula is: \[s_{d}^{2} = \frac{\sum_{i=1}^{n} (d_i - \bar{d})^2}{n-1}\].\ This involves computing how each difference \(d_i\) deviates from the mean difference, squaring these deviations, summing them up, and then dividing by \(n-1\), the number of subjects minus one. In the given exercise, the variance \(s_{d}^{2}\) was calculated to be 0.132. This tells us that the dispersion of temperature differences in our sample is relatively low but gives more precise information on how spread out the differences are.
statistical interpretation
Statistical interpretation involves understanding and drawing conclusions from the calculated values. In this case, the mean difference \(\bar{d} = 0.28\) indicates that, on average, body temperatures are slightly higher at 12 AM compared to 8 AM. The standard deviation \(s_d = 0.36\) shows that while individual differences in body temperature vary, they are centered around the mean difference. Lastly, the variance \(s_{d}^{2} = 0.132\) represents the degree of spread in these differences. By interpreting these values, one can better understand whether the observed differences are significant or if they might fall within expected random variations.
paired data analysis
Paired data analysis is a statistical approach used when comparing two related sets of data. It focuses on the differences within pairs rather than treating them as independent subjects. This method is crucial when the same subjects are measured twice under different conditions, like in our exercise measuring body temperatures at different times of the day. By analyzing the differences, we can remove subject-specific variability. The steps in paired data analysis typically include:
  • Calculating the differences for each pair.
  • Computing the mean difference.
  • Finding the sample variance and standard deviation.
This method provides a more precise understanding of the changes between conditions than analyzing the measurements separately.

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Most popular questions from this chapter

Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9 - 1 along with "Table" answers based on Table \(A\) - 3 with df equal to the smaller of \(\boldsymbol{n}_{I}-\boldsymbol{I}\) and \(\boldsymbol{n}_{2}-\boldsymbol{I} .\) ) Blanking Out on Tests Many students have had the unpleasant experience of panicking on a test because the first question was exceptionally difficult. The arrangement of test items was studied for its effect on anxiety. The following scores are measures of "debilitating test anxiety." which most of us call panic or blanking out (based on data from "Item Arrangement, Cognitive Entry Characteristics, Sex and Test Anxiety as Predictors of Achievement in Examination Performance," by Klimko, Journal of Experimental Education, Vol. 52, No. 4.) Is there sufficient evidence to support the claim that the two populations of scores have different means? Is there sufficient evidence to support the claim that the arrangement of the test items has an effect on the score? Is the conclusion affected by whether the significance level is 0.05 or 0.01? $$\begin{array}{|c|c|c|c|c|} \hline {}{} {\text { Questions Arranged from Easy to Difficult }} \\ \hline 24.64 & 39.29 & 16.32 & 32.83 & 28.02 \\ \hline 33.31 & 20.60 & 21.13 & 26.69 & 28.90 \\ \hline 26.43 & 24.23 & 7.10 & 32.86 & 21.06 \\ \hline 28.89 & 28.71 & 31.73 & 30.02 & 21.96 \\ \hline 25.49 & 38.81 & 27.85 & 30.29 & 30.72 \\ \hline\end{array}$$ $$\begin{array}{|c|c|c|c|} \hline {}{} {\text { Questions Arranged from Difficult to Easy }} \\ \hline 33.62 & 34.02 & 26.63 & 30.26 \\ \hline 35.91 & 26.68 & 29.49 & 35.32 \\ \hline 27.24 & 32.34 & 29.34 & 33.53 \\ \hline 27.62 & 42.91 & 30.20 & 32.54 \\ \hline\end{array}$$

In the article "On Judging the Significance of Differences by Examining the Overlap Between Confidence Intervals," by Schenker and Gentleman (American Statistician, Vol. 55, No. 3), the authors consider sample data in this statement: "Independent simple random samples, each of size \(200\) , have been drawn, and 112 people in the first sample have the attribute, whereas 88 people in the second sample have the attribute." a. Use the methods of this section to construct a \(95 \%\) confidence interval estimate of the difference \(p_{1}-p_{2} .\) What does the result suggest about the equality of \(p_{1}\) and \(p_{2} ?\) b. Use the methods of Section \(7-1\) to construct individual \(95 \%\) confidence interval estimates for each of the two population proportions. After comparing the overlap between the two confidence intervals, what do you conclude about the equality of \(p_{1}\) and \(p_{2} ?\) c. Use a 0.05 significance level to test the claim that the two population proportions are equal. What do you conclude? d. Based on the preceding results, what should you conclude about the equality of \(p_{1}\) and \(p_{2} ?\) Which of the three preceding methods is least effective in testing for the equality of \(p_{1}\) and \(p_{2} ?\)

Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. In a study of Burger King drive-through orders, it was found that 264 orders were accurate and 54 were not accurate. For McDonald's, 329 orders were found to be accurate while 33 orders were not accurate (based on data from \(Q S R\) magazine). Use a 0.05 significance level to test the claim that Burger King and McDonald's have the same accuracy rates. a. Test the claim using a hypothesis test. b. Test the claim by constructing an appropriate confidence interval. c. Relative to accuracy of orders, does either restaurant chain appear to be better?

Use the listed paired sample data, and assume that the samples are simple random samples and that the differences have a distribution that is approximately normal. A popular theory is that presidential candidates have an advantage if they are taller than their main opponents. Listed are heights (cm) of presidents along with the heights of their main opponents (from Data Set 15 "Presidents"). a. Use the sample data with a 0.05 significance level to test the claim that for the population of heights of presidents and their main opponents, the differences have a mean greater than \(0 \mathrm{cm} .\) b. Construct the confidence interval that could be used for the hypothesis test described in part (a). What feature of the confidence interval leads to the same conclusion reached in part (a)? $$\begin{array}{|l|l|l|l|l|l|l|} \hline \text { Height (cm) of President } & 185 & 178 & 175 & 183 & 193 & 173 \\\ \hline \text { Height (cm) of Main Opponent } & 171 & 180 & 173 & 175 & 188 & 178 \\ \hline \end{array}$$

Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9 - 1 along with "Table" answers based on Table \(A\) - 3 with df equal to the smaller of \(\boldsymbol{n}_{I}-\boldsymbol{I}\) and \(\boldsymbol{n}_{2}-\boldsymbol{I} .\) ) Color and Cognition Researchers from the University of British Columbia conducted a study to investigate the effects of color on cognitive tasks. Words were displayed on a computer screen with background colors of red and blue. Results from scores on a test of word recall are given below. Higher scores correspond to greater word recall. a. Use a 0.05 significance level to test the claim that the samples are from populations with the same mean. b. Construct a confidence interval appropriate for the hypothesis test in part (a). What is it about the confidence interval that causes us to reach the same conclusion from part (a)? c. Does the background color appear to have an effect on word recall scores? If so, which color appears to be associated with higher word memory recall scores? $$\begin{array}{l|l} \hline \text { Red Background } & n=35, \mathrm{x}=15.89, \mathrm{s}=5.90 \\ \hline \text { Blue Background } & n=36, \mathrm{x}=12.31, \mathrm{s}=5.48 \\ \hline \end{array}$$

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