Question

Short answer

The 99% confidence interval estimate of the mean of the differences between brain volumes for the first-born and the second-born is equal to(-66.75\({\rm{c}}{{\rm{m}}^3}\),49.75\({\rm{c}}{{\rm{m}}^3}\)).

As the value of 0 lies in the interval,there is a possibility that the brain volume of the firstborn is equal to the brain volume of the second born.

Step-by-step solution

Given information

The data are given on the brain volumes (in \({\rm{c}}{{\rm{m}}^{^3}}\)) of twins.

Hypotheses

Null Hypothesis: The mean of the differences between the brain volumes is equal to 0.

\({H_{0\;}}:\;{\mu _d} = 0\)

Alternative Hypothesis:The mean of the differences between the brain volumes is not equal to 0.\({H_1}\;:{\mu _d} \ne 0\;\..

Here \({\mu _d}\;\)is mean of the population of differences between the brain volumes of the firstborn and the second born.

Step 3:Expression of the confidence interval

The formula of the confidence interval is as follows:

\({\rm{C}}{\rm{.I}} = \bar d - E < {\mu _d} < \bar d + E\;\)

The value of the margin of error € has the following notation:

\(E = {t_{\frac{\alpha }{2},df}} \times \frac{{{s_d}}}{{\sqrt n }}\)

Table of the differences

First Born	Second Born	Differences (d)
1005	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

Find the mean of the differences

The mean of the differences of the samples is equal to:

\(\begin{array}{c}\bar d = \frac{{\sum\limits_{i = 1}^n {{d_i}} }}{n}\\ = \frac{{42 + 8 + ... + \left( { - 44} \right)}}{{10}}\\ = - 8.5\end{array}\)

Find the standard deviation of the differences

The standard deviation of thedifferences is equal to:

\(\begin{array}{c}s = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{\left( {{d_i} - \bar d} \right)}^2}} }}{{n - 1}}} \;\\ = \sqrt {\frac{{{{\left( {42 - \left( { - 8.5} \right)} \right)}^2} + {{\left( {8 - \left( { - 8.5} \right)} \right)}^2} + \cdots + {{\left( {\left( { - 44} \right) - \left( { - 8.5} \right)} \right)}^2}}}{{10 - 1}}} \\ = 56.68\end{array}\)

Find the critical value

The degrees of freedom will be equal to:

\(\begin{array}{c}{\rm{df}} = n - 1\\ = 10 - 1\\ = 9\end{array}\)

The confidence level is equal to 99%. Thus, the level of significance becomes 0.01.

Therefore,

\(\begin{array}{c}\frac{\alpha }{2} = \frac{{0.01}}{2}\\ = 0.005\end{array}\)

Referring to the t-distribution table, the critical value of t for 9 degrees of freedom at 0.005 significance level is equal to 3.2498.

Compute the margin of error

The value of the margin of error is computed below:

\(\begin{array}{c}E = {t_{\frac{\alpha }{2},df}} \times \frac{{{s_d}}}{{\sqrt n }}\\ = {t_{\frac{{0.01}}{2},9}} \times \frac{{56.68}}{{\sqrt {10} }}\\ = 3.2498 \times \frac{{56.68}}{{\sqrt {10} }}\\ = 58.24762\end{array}\)

Step 9:Compute the confidence interval

The value of the confidence interval is equal to:

\(\begin{array}{c}\bar d - E < {\mu _d} < \bar d + E\;\\\left( { - 8.5 - 58.24762} \right) < {\mu _d} < \left( { - 8.5 + 58.24762} \right)\\ - 66.75 < {\mu _d} < 49.75\end{array}\)

Thus, the 99% confidence interval is equal to (-66.75,49.75).

Interpretation of the confidence interval

It can be seen that the value of 0 is included in the interval.

This implies that the mean of the differences between the brain volumes can be equal to 0.

Thus, there is a possibility that the brain volume of the firstborn is equal to the brain volume of the second born.