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

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.

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 鈧� has the following notation:

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

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

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

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

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

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

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.

The margin of error is equal to:

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

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).

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.