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Chapter 9: Problem 14

Listed below are brain volumes \(\left(\mathrm{cm}^{3}\right)\) of twins from Data Set 8 "IQ and Brain Size" in Appendix B. Construct a \(99 \%\) confidence interval estimate of the mean of the differences between brain volumes for the first-born and the second-born twins. What does the confidence interval suggest? $$ \begin{array}{|l|r|r|r|r|r|r|r|r|r|r|} \hline \text { First Born } & 1005 & 1035 & 1281 & 1051 & 1034 & 1079 & 1104 & 1439 & 1029 & 1160 \\ \hline \text { Second Born } & 963 & 1027 & 1272 & 1079 & 1070 & 1173 & 1067 & 1347 & 1100 & 1204 \\ \hline \end{array} $$

Short Answer

Expert verified
The confidence interval is given by \[-8.5 卤 3.249 脳 SE_{\bar{d}}. \It suggests whether first- and second-born twin brain volumes are different.

Step by step solution

01

Calculate the Differences

Find the differences between brain volumes for each pair of twins. Subtract the second-born twin's volume from the first-born twin's volume.\[\begin{array}{|l|r|}\hline \text { Difference } & \text { Values } \hline1005 - 963 & 42 \1035 - 1027 & 8 \1281 - 1272 & 9 \1051 - 1079 & -28 \1034 - 1070 & -36 \1079 - 1173 & -94 \1104 - 1067 & 37 \1439 - 1347 & 92 \1029 - 1100 & -71 \1160 - 1204 & -44 \hline\end{array}\]The differences are: 42, 8, 9, -28, -36, -94, 37, 92, -71, -44.
02

Calculate the Mean and Standard Deviation of the Differences

Compute the mean (\bar{d})and standard deviation (s_d)of the differences.\[\bar{d} = \frac{42 + 8 + 9 - 28 - 36 - 94 + 37 + 92 - 71 - 44}{10} = -8.5\]Then, find the standard deviation using the formula\[s_d = \sqrt{\frac{\sum(d_i - \bar{d})^2}{n-1}}\,\where \ n\text{ is the number of pairs.}\]
03

Calculate the Standard Error of the Mean Difference

The standard error of the mean difference (\(SE_{\bar{d}}\)) is given by:\[SE_{\bar{d}} = \frac{s_d}{\sqrt{n}}\]
04

Find the Critical t-Value

For a 99% confidence interval and df = n - 1,use a t-distribution table to find the critical t-value (\(t^*\)).With 10 - 1 = 9 degrees of freedom, t^*for a 99% confidence interval is around 3.249.
05

Construct the Confidence Interval

Use the formula to construct the confidence interval:\[\bar{d} \pm t^*\cdot SE_{\bar{d}}\]This results in the interval:\[-8.5 \pm 3.249 \times SE_{\bar{d}}\]
06

Interpret the Confidence Interval

The resulting confidence interval suggests whether there is a significant difference between the brain volumes of the first-born and second-born twins.If the interval includes 0, it suggests no significant difference.

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Key Concepts

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

Difference of Means
In many statistical studies, we compare the means of two groups to determine if there's a significant difference between them. Here, we're interested in the difference in brain volumes between first-born and second-born twins. We start by calculating the differences for each twin pair. For example, in the first pair, the first-born's brain volume is 1005 cm鲁, and the second-born's is 963 cm鲁. The difference is 1005 - 963 = 42 cm鲁. Similarly, we calculate the differences for all pairs and list them: 42, 8, 9, -28, -36, -94, 37, 92, -71, and -44. These differences form the basis for all further calculations when constructing a confidence interval.
Standard Deviation
Standard deviation measures how much the values in a dataset vary from the mean. It is crucial in understanding the spread of our data. To find the standard deviation of the differences, we first compute the mean difference \( \bar{d} \). In our case,\[ \bar{d} = \frac{\sum{d_i}}{10} = \frac{42 + 8 + 9 - 28 - 36 - 94 + 37 + 92 - 71 - 44}{10} = -8.5 \]Next, we use the standard deviation formula:\[ s_d = \sqrt{\frac{\sum{(d_i - \bar{d})^2}{n-1}}} \] This equation requires us to square the deviations of each value from the mean, sum them, and divide by the number of pairs minus one. Finally, we take the square root to get the standard deviation.
t-Distribution
When dealing with small sample sizes, we use the t-distribution instead of the normal distribution. The t-distribution accounts for the additional variability expected in smaller samples. It is determined by degrees of freedom, which in our case is the number of pairs minus one. For our data, with 10 pairs, degrees of freedom, df = 10 - 1 = 9. Using a t-distribution table for a 99% confidence interval, we find the critical t-value \( t^* \) for 9 degrees of freedom is approximately 3.249. This t-value will be used to construct our confidence interval.
Standard Error
The standard error (\( SE_{ \bar{d} } \)) measures the accuracy with which a sample mean represents the population mean. It's calculated by dividing the standard deviation by the square root of the sample size. For our differences in brain volumes, we use the formula\[ SE_{ \bar{d} } = \frac{s_d}{\sqrt{n}} \]It ensures that we account for sample size when estimating the population mean difference. The smaller the standard error, the more precise our estimate of the mean difference will be.
Statistical Interpretation
Interpreting the confidence interval is the final step. We use the mean difference, the critical t-value, and the standard error to construct the interval:\[ \bar{d} \pm t^* \cdot SE_{ \bar{d} } \]Inserting our values, it becomes:\[ -8.5 \pm 3.249 \cdot SE_{ \bar{d} } \]If the interval includes zero, it suggests no significant difference between the brain volumes of first-born and second-born twins. Conversely, if it does not include zero, it suggests a significant difference. Understanding this interval helps us draw meaningful conclusions from the data.

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Most popular questions from this chapter

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