Question

Short answer

The 99% estimate of confidence interval of the mean difference between reported heights and measured heights is equal to (-4.16 inches,2.16 inches).

The confidence interval shows that there is some difference between reported heights and measured heights. Also, the confidence limits contain 0. Thus, the reported heights and the measured heights can also be equal to 0.

Step-by-step solution

Given information

The given data consists of the self-reported heights and the measured heights of males aged 12-16.

Hypotheses

Null Hypothesis: The mean of the differences between the reported heights and measured heights is equal to 0.

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

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

\({\mu _d}\) is the mean of the population of difference between the reported heights and measured heights.

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

The following table shows the difference between the reported heights and the measured heights:

Reported	68	71	63	70	71	60	65	64	54	63	66	72
Measured	67.9	69.9	64.9	68.3	70.3	60.6	64.5	67	55.6	74.2	65	70.8
Differences(d)	0.1	1.1	-1.9	1.7	0.7	-0.6	0.5	-3	-1.6	-11.2	1	1.2

Find the mean of the differences

Now, the mean of the differences between the reported and measured heights is computed below:

\(\begin{array}{c}\bar d = \frac{{\sum\limits_{i = 1}^n {{d_i}} }}{n}\\ = \frac{{0.1 + 1.1 + \cdots + 1.2}}{{12}}\\ = - 1\end{array}\)

Find the standard deviation

The standard deviation of the differences between the reported and the measured heights is computed below:

\(\begin{array}{c}s = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{\left( {{d_i} - \bar d} \right)}^2}} }}{{n - 1}}} \;\\ = \sqrt {\frac{{{{\left( {0.1 - \left( { - 1} \right)} \right)}^2} + {{\left( {1.1 - \left( { - 1} \right)} \right)}^2} + \cdots + {{\left( {1.2 - \left( { - 1} \right)} \right)}^2}}}{{12 - 1}}} \\ = 3.520\end{array}\)

Find the critical value

The degrees of freedom are computed as follows:

\(\begin{array}{c}{\rm{df}} = n - 1\\ = 12 - 1\\ = 11\end{array}\)

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 11 degrees of freedom at 0.005 significance level is equal to 3.1058.

Compute the margin of error

The margin of error is equal to:

\(\begin{array}{c}E = {t_{\frac{\alpha }{2},df}} \times \frac{s}{{\sqrt n }}\\ = {t_{\frac{{0.01}}{2},11}} \times \frac{{3.520}}{{\sqrt {12} }}\\ = 3.1058 \times \frac{{3.520}}{{\sqrt {12} }}\\ = 3.155518\end{array}\)

Step 9:Compute the confidence interval

The confidence interval is computed as follows:

\(\begin{array}{c}\bar d - E < {\mu _d} < \bar d + E\;;\\\left( { - 1 - 3.155518} \right) < {\mu _d} < \left( { - 1 + 3.155518} \right)\\ - 4.16 < {\mu _d} < 2.16\end{array}\)

Thus, the 99% confidence interval is equal to (-4.16 inches,2.16 inches).

Interpretation of the confidence interval

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

This implies that the differences between the reported heights and the measured heights can be equal to 0.

Thus, there is not enough evidence to conclude that thereported heights and measured heights are not equal.