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Chapter 9: Problem 16

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. As part of the National Health and Nutrition Examination Survey, the Department of Health and Human Services obtained self-reported heights (in.) and measured heights (in.) for males aged \(12-16 .\) Listed below are sample results. Construct a \(99 \%\) confidence interval estimate of the mean difference between reported heights and measured heights. Interpret the resulting confidence interval, and comment on the implications of whether the confidence interval limits contain \(0 .\) $$\begin{array}{|l|c|c|c|c|c|c|c|c|c|c|c|c|} \hline \text { Reported } & 68 & 71 & 63 & 70 & 71 & 60 & 65 & 64 & 54 & 63 & 66 & 72 \\ \hline \text { Measured } & 67.9 & 69.9 & 64.9 & 68.3 & 70.3 & 60.6 & 64.5 & 67.0 & 55.6 & 74.2 & 65.0 & 70.8 \\ \hline \end{array}$$

Short Answer

Expert verified
Calculate mean difference and standard deviation of differences. Find the margin of error and create the confidence interval to interpret the result.

Step by step solution

01

- Calculate the Differences

Calculate the difference between each pair of reported and measured heights. For example, for the first pair, the difference is: \[ 68 - 67.9 = 0.1 \]
02

- Find the Mean Difference

Sum all the differences calculated in Step 1 and then divide by the number of pairs (n=12) to find the mean difference (\( \bar{d} \)).
03

- Find the Standard Deviation of Differences

Use the differences calculated in Step 1 to find the standard deviation (\( s_d \)) of the differences. First, calculate the variance by finding the average squared differences from the mean. Then, take the square root of the variance to get the standard deviation.
04

- Calculate the Standard Error

The standard error (SE) of the mean difference is calculated using the formula: \[ SE = \frac{s_d}{\sqrt{n}} \] where \( s_d \) is the standard deviation of the differences and \( n \) is the number of observations (12).
05

- Find the Critical Value

For a 99% confidence interval with 11 degrees of freedom (n-1=12-1), find the critical t-value (\( t_{\alpha/2} \)) using a t-distribution table.
06

- Calculate the Margin of Error

The margin of error (ME) is calculated using the formula: \[ ME = t_{\alpha/2} \cdot SE \]
07

- Construct the Confidence Interval

Use the mean difference and margin of error to create the confidence interval: \[ \bar{d} \pm ME \] Interpret the interval in the context of the problem.
08

- Interpret the Confidence Interval

Evaluate if 0 is contained within the confidence interval. If 0 is within the interval, it implies no significant difference between reported and measured heights. Comment on the results.

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Key Concepts

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

Paired Sample Data
Paired sample data refer to data sets where every measurement in one sample is uniquely paired with a measurement in the second sample. In the problem, each reported height is linked to a corresponding measured height for each male aged 12-16. This direct pairing allows us to directly compare the two sets of data.

We analyze the differences between paired measurements rather than treating each group separately. For example, with reported and measured heights, the first pair is 68 inches (reported) and 67.9 inches (measured). We calculate the difference for each pair to determine if there's a consistent deviation between self-reported and actual heights.
Mean Difference
The mean difference (\( \bar{d} \)) is the average of all the differences calculated from the paired sample data. First, find each difference by subtracting measured height from reported height, then sum all these differences, and divide by the total number of pairs (n).

In our exercise, after calculating the difference for each pair of heights, we sum them up and divide by 12 (since there are 12 pairs). This mean difference helps us understand the average deviation between reported and measured heights, providing a central value for further analysis.
Standard Deviation of Differences
The standard deviation of differences (\( s_d \)) quantifies how much the differences vary around the mean difference. To find this:

  • First, calculate each difference from the mean difference.
  • Square these deviations.
  • Find the average of these squared deviations (this is the variance).
  • Take the square root of the variance to get the standard deviation of differences.


This measure helps us determine how spread out the differences are; a smaller standard deviation indicates that the differences are closer to the mean difference, while a larger one shows more variability.
t-Distribution
The t-distribution is used when constructing confidence intervals for small sample sizes (typically n < 30) or when the population standard deviation is unknown. It accounts for the additional variability introduced by estimating the population standard deviation from a sample.

For our confidence interval, we use the t-distribution with degrees of freedom (df) equal to n-1. Here, df = 12-1 = 11. The critical value (\( t_{\frac{\alpha}{2}} \)) is obtained based on this df and the desired confidence level (99%). This value helps determine the margin of error and thus the bounds of our confidence interval.
Margin of Error
The margin of error (ME) represents the range within which we expect the true mean difference to lie, with a given level of confidence. It is calculated by multiplying the critical t-value (\( t_{\frac{\alpha}{2}} \)) by the standard error.

The standard error (SE) is found using: \[ SE = \frac{s_d}{\sqrt{n}} \]
Once we have SE and the critical t-value, the margin of error can be calculated as: \[ ME = t_{\frac{\alpha}{2}} \cdot SE \]
This ME is then added to and subtracted from the mean difference to form the confidence interval (\( \bar{d} \pm ME \)). This interval tells us, with 99% confidence, where the true mean difference between reported and measured heights lies.

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Most popular questions from this chapter

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. Listed below are body temperatures from seven different subjects measured at two different times in a day (from Data Set 3 "Body Temperatures" in Appendix B). a. Use a 0.05 significance level to test the claim that there is no difference between body temperatures measured at \(8 \mathrm{AM}\) and at \(12 \mathrm{AM}\). 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)? $$\begin{array}{l|l|l|l|l|l|l|l} \hline \text { Body Temperature ("F) at 8 AM } & 96.6 & 97.0 & 97.0 & 97.8 & 97.0 & 97.4 & 96.6 \\ \hline \text { Body Temperature ('F) at 12 AM } & 99.0 & 98.4 & 98.0 & 98.6 & 98.5 & 98.9 & 98.4 \\ \hline \end{array}$$

In the largest clinical trial ever conducted, 401,974 children were randomly assigned to two groups. The treatment group consisted of 201,229 children given the Salk vaccine for polio, and 33 of those children developed polio. The other 200,745 children were given a placebo, and 115 of those children developed polio. If we want to use the methods of this section to test the claim that the rate of polio is less for children given the Salk vaccine, are the requirements for a hypothesis test satisfied? Explain.

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)? $$\begin{array}{|l|l|l|l|l|l|l|} \hline \text { Height (cm) of President } & 185 & 178 & 175 & 183 & 193 & 173 \\\ \hline \text { Height (cm) of Main Opponent } & 171 & 180 & 173 & 175 & 188 & 178 \\ \hline \end{array}$$

Confidence Interval for Hemoglobin Large samples of women and men are obtained, and the hemoglobin level is measured in each subject. Here is the \(95 \%\) confidence interval for the difference between the two population means, where the measures from women correspond to population 1 and the measures from men correspond to population 2 : \(-1.76 \mathrm{g} / \mathrm{dL}<\mu_{1}-\mu_{2}<-1.62 \mathrm{g} / \mathrm{dL}\) a. What does the confidence interval suggest about equality of the mean hemoglobin level in women and the mean hemoglobin level in men? b. Write a brief statement that interprets that confidence interval. c. Express the confidence interval with measures from men being population 1 and measures from women being population 2

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 \(\boldsymbol{n}_{2}-\boldsymbol{I} .\) ) Is Old Faithful Not Quite So Faithful? Listed below are time intervals (min) between eruptions of the Old Faithful geyser. The "recent" times are within the past few years, and the "past" times are from 1995. Does it appear that the mean time interval has changed? Is the conclusion affected by whether the significance level is 0.05 or \(0.01 ?\) $$\begin{array}{l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l} \hline \text { Recent } & 78 & 91 & 89 & 79 & 57 & 100 & 62 & 87 & 70 & 88 & 82 & 83 & 56 & 81 & 74 & 102 & 61 \\ \hline \text { Past } & 89 & 88 & 97 & 98 & 64 & 85 & 85 & 96 & 87 & 95 & 90 & 95 & & & & & \\ \hline\end{array}$$
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