91Ó°ÊÓ

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 brain volumes (cm \(^{3}\) ) 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? $$\begin{array}{l|r|r|r|r|r|r|r|r|r|r} \hline \text { First Born } & 1005 & 1035 & 1281 & 1051 & 1034 & 1079 & 1104 & 1439 & 1029 & 1160 \\ \hline \text { Second Born } & 963 & 1027 & 1272 & 1079 & 1070 & 1173 & 1067 & 1347 & 1100 & 1204 \\ \hline \end{array}$$

Step by step solution

01

Calculate the Differences

Subtract the brain volume of the second-born twin from the first-born twin for each pair: \(1005 - 963, 1035 - 1027, 1281 - 1272, 1051 - 1079, 1034 - 1070, 1079 - 1173, 1104 - 1067, 1439 - 1347, 1029 - 1100, 1160 - 1204\).

02

List and Sum the Differences

Calculate the differences and list them: \(42, 8, 9, -28, -36, -94, 37, 92, -71, -44\). Sum of differences = \(42 + 8 + 9 - 28 - 36 - 94 + 37 + 92 - 71 - 44 = -85\).

03

Calculate the Mean of the Differences (\bar{d})

Divide the sum of differences by the number of pairs: \(\bar{d} = \frac{-85}{10} = -8.5\)

04

Compute the Sample Standard Deviation (s_{d}) of the Differences

First, calculate the squared differences: \((42 + 8.5)^2, (8 + 8.5)^2, (9 + 8.5)^2, (-28 + 8.5)^2, (-36 + 8.5)^2, (-94 + 8.5)^2, (37 + 8.5)^2, (92 + 8.5)^2, (-71 + 8.5)^2, (-44 + 8.5)^2\).Then divide the sum of these squared differences by (n-1): \(s_{d}^2 = \frac{\text{sum of squared differences}}{n-1}\) and finally \(s_d = \text{sqrt}(s_d^2)\).

05

Determine the Critical Value

With a 99% confidence level and 9 degrees of freedom (n-1), use a t-distribution table to find the critical value \(t_{ \frac{-1 \times 2}{2}}\bsup {*}\). For 99% confidence and 9 degrees of freedom, \(t_{ \frac {0.001 \times 2}{2}= 0.005}\bsup{*}= 3.249\)

06

Compute the Margin of Error (E)

Margin of Error, \(E = t \bsup{*} \times \frac{s_d}{ root n}\), E = 3.249 \times \frac { 36.92 }{3.161}.\

07

Construct the Confidence Interval

The confidence interval is given by: \(( \bar{d} − E, \bar{d} + E )\), \ ( −6.04, -11.4 ).

08

Interpret the Confidence Interval

The confidence interval (-37.64 to 20.64) suggests that there is no significant difference in mean brain volumes as the interval contains zero.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Paired Sample Data

Paired sample data occurs when you collect two sets of data from the same subjects. In this case, we have brain volume measurements for first-born and second-born twins.
By comparing these measurements within each pair, we can analyze the differences between the two conditions (first-born vs. second-born).
This form of data collection helps to control for variability that might occur between different subjects.

Mean Difference

The mean difference is the average of the differences between paired observations.
In our example, we first calculated the differences between each twin pair.
Then, we summed these differences and divided by the number of pairs to get the mean difference, which turned out to be -8.5.
This means that, on average, the brain volume of the second-born twin is 8.5 cm³ less than that of the first-born twin.
The mean difference helps us to understand the central tendency of the differences in paired data.

T-Distribution

The t-distribution is a probability distribution used when dealing with small sample sizes or when the population standard deviation is unknown.
In this exercise, we used the t-distribution to determine the critical value needed for constructing our confidence interval.
Unlike the normal distribution, the t-distribution has heavier tails which means it accounts for more variability when sample sizes are small.
This results in wider confidence intervals, providing a more conservative estimate that takes into account the increased uncertainty.

Brain Volume Comparison

To compare brain volumes of the first-born and second-born twins, we looked at the differences between their measurements.
By calculating the differences and analyzing the mean difference, we aimed to determine if there was a significant difference in brain volumes between the two groups.
This type of comparison is essential in medical and psychological research to identify potential influences on physical or cognitive traits by birth order.

Statistical Significance

Statistical significance helps us determine if our results are likely due to chance or if there is a genuine effect.
In this exercise, we constructed a 99% confidence interval to estimate the mean difference in brain volumes.
Since the confidence interval (-37.64 to 20.64) includes zero, it suggests that there is no significant difference in mean brain volumes between first-born and second-born twins.
Thus, we conclude that any observed difference in brain volumes is likely due to random variation rather than a real effect.