Chapter 24, Problem 14E

(a)

To determine

To explain are these data appropriate for inference.

Answer to Problem 14E

Yes, it is appropriate.

Explanation of Solution

For this data, we need to check that this data is appropriate for inference or not. So, we have,

Random condition: It is satisfied because assuming that the skulls were randomly selected.

Independent condition: It is satisfied because the individuals in $4000$ B.C. are independent of the individuals in $200$ B.C.E.

Normal condition: It is satisfied because both sample sizes of $30$ are at least $30$ and the central limit theorem then tells us that the sampling distribution of the sample mean is approximately normal.

Thus, these data is appropriate for inference.

(b)

To determine

To create $95\%$ confidence interval for the difference in mean skull breadth between these two eras.

Answer to Problem 14E

The confidence interval is: $(-6.6611,-1.8721)$ .

Explanation of Solution

It is given that:

Thus, by calculating the mean and standard deviation of the data using excel function we get,

The formulas used in excel to find the mean and standard deviations are as follows:

1. AVERAGE ( $\text{number1, }\left[\text{number2}\right]$, ...): Returns the average (arithmetic mean) of the arguments.
2. STDEV ( $\text{number1, }\left[\text{number2}\right]$,...): Estimates standard deviation based on a sample.

For the $4000$ B.C.:

4000 B.C.E.
Average	=AVERAGE(C1:C30)
S.D.	=STDEV(C1:C30)

And the result will be as:

4000 B.C.E.
Average	131.3667
S.D.	5.1292

Now, for the $200$ B.C.E.:

200 B.C.E.
Average	=AVERAGE(D1:D30)
S.D.	=STDEV(D1:D30)

And the result will be:

200 B.C.E.
Average	135.6333
S.D.	4.0385

Thus, we have,

  $\begin{array}{l}{\overline{x}}_1=131.3667\\ {\overline{x}}_2=135.6333\\ n_1=30\\ n_2=30\\ s_1=5.1292\\ s_2=4.0385\\ \alpha =0.05\end{array}$

Thus, the degree of freedom will be:

  $df=54$

And the t -value is as:

  $t=2.009$

Thus, the confidence interval is:

  $\begin{array}{l}({\overline{x}}_1-{\overline{x}}_2)-t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =(131.3667-135.6333)-2.009\times \sqrt{\frac{{5.1292}^2}{30}+\frac{{4.0385}^2}{30}}\\ =-6.6611\\ ({\overline{x}}_1-{\overline{x}}_2)+t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =(131.3667-135.6333)+2.009\times \sqrt{\frac{{5.1292}^2}{30}+\frac{{4.0385}^2}{30}}\\ =-1.8721\end{array}$

(c)

To determine

To explain do these data provide evidence that the mean breadth of men’s skulls changed over these period.

Answer to Problem 14E

Yes.

Explanation of Solution

Thus, we have,

  $\begin{array}{l}{\overline{x}}_1=131.3667\\ {\overline{x}}_2=135.6333\\ n_1=30\\ n_2=30\\ s_1=5.1292\\ s_2=4.0385\\ \alpha =0.05\end{array}$

Thus, the confidence interval is: $(-6.6611,-1.8721)$ . Yes, these data provide evidence that the mean breadth of men’s skulls changed over these period since the confidence interval does not contain zero, this indicates that the mean of the two groups is significantly different.