Chapter 11.1, Problem 11E

Are you smarter than your older brother? In a study of birth order and intelligence, IQ tests were given to 18- and 19-year-old men to estimate the size of the difference, if any, between the mean IQs of firstborn sons and secondborn sons. The following data for 10 firstborn sons and 10 secondborn sons are consistent with the means and standard deviations reported in the article. It is reasonable to assume that the samples come from populations that are approximately normal.

Can you conclude that there is a difference in mean IQ between firstborn and secondborn sons?

Use the $\alpha =0.01$ level.

To determine

To find: Whether it could be concluded that there is difference in mean IQ between firstborn and second born sons.

Answer to Problem 11E

No, it cannot be concludedthat there is difference in mean IQ between firstborn and second born sons.

Explanation of Solution

Given:

First born	Second born
104	103
82	103
102	91
96	113
129	102
89	103
114	92
107	90
89	114
129	113

The provided table can be tabulated as:

First born	Second born	Difference
104	103	1
82	103	-21
102	91	11
96	113	-17
129	102	27
89	103	-14
114	92	22
107	90	17
89	114	-25
129	113	16

The null and alternative hypotheses are:

  $H_0:{\mu }_d=0$

That is, there is no difference in mean IQ between firstborn and second born sons.

  $H_a:{\mu }_d\ne 0$

That is, there is a difference in mean IQ between firstborn and second born sons.

The alternative hypothesis suggests that the test is two-tailed

It is given that the significance level is $\alpha =0.01$ .

Degree of freedom can be calculated as:

  $\begin{array}{c}\text{Degree of freedom}=n-1\\ =10-1\\ =9\end{array}$

The critical value at 1% significance level and degree of freedom = 94 is:

  $\begin{array}{c}{t}_{\alpha /2,df}=t_{0.01/2,9}\\ =3.250\text{ (Using t-table)}\end{array}$

To compute the test statistics values of sample mean, and sample standard deviation of differences are required which can be calculated as:

  $\begin{array}{c}\overline{d}=\frac{{\displaystyle \sum _{i=1}^n{d}_i}}{n}\\ =\frac{1+\left(-21\right)+\mathrm{...}+16}{10}\\ =1.7\\ s_d=\sqrt{\frac{{\displaystyle \sum _{i=1}^n{\left(d_i-\overline{d}\right)}^2}}{n-1}}\\ =\sqrt{\frac{{\left(1-1.7\right)}^2+{\left(-21-1.7\right)}^2+\mathrm{...}+{\left(16-16\right)}^2}{10-1}}\\ =19.442\end{array}$

The test statistic can be computed as:

  $\begin{array}{c}t=\frac{\overline{d}-{\mu }_d}{\frac{s_d}{\sqrt{n}}}\\ =\frac{1.7-0}{\frac{19.442}{\sqrt{10}}}\\ =0.277\end{array}$

Decision rule (for two tail test): Reject null hypothesis, if $\left|t\right|>t_{\alpha ,df}$

Here, test statistic = 0.277, that is, |t| = 0.277.

So, |t| <3.250, the correct decision is not to reject $H_0$

Thus, it could be said that there is insufficient evidence to conclude that there is difference in mean IQ between firstborn and second born sons.