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Notice first that the true percentage differences needs to be tested; thus, first find all 10 percentage differences

- for every meal. The percentage can be computed using

\({x_i}{\rm{ = }}\frac{{{\rm{ Measured}}{{\rm{ }}_{\rm{i}}}{\rm{ - Stated}}{{\rm{ }}_{\rm{i}}}}}{{{\rm{ Stated}}{{\rm{ }}_{\rm{i}}}}},\quad i = 1,2, \ldots ,10.\)

\(\begin{array}{l}{x_1} = \frac{{212 - 180}}{{180}} = 0.1778 \approx 17.78\% ;\\{x_2} = \frac{{319 - 220}}{{220}} = 0.45 \approx 45\% ;\\{x_3} = \frac{{231 - 190}}{{190}} = 0.2158 \approx 21.58\% ;\\{x_4} = \frac{{306 - 230}}{{230}} = 0.3304 \approx 33.04\% ;\\{x_5} = \frac{{211 - 200}}{{200}} = 0.055 \approx 5.5\% ;\\{x_6} = \frac{{431 - 370}}{{370}} = 0.1649 \approx 16.49\% ;\\{x_7} = \frac{{288 - 250}}{{250}} = 0.152 \approx 15.2\% ;\\{x_8} = \frac{{265 - 240}}{{240}} = 0.1042 \approx 10.42\% ;\\{x_9} = \frac{{145 - 80}}{{80}} = 0.8125 \approx 81.25\% ;\\{x_{10}} = \frac{{228 - 180}}{{180}} = 0.2667 \approx 26.67\% ;\end{array}\)

The hypotheses of interest are\({H_0}:\mu = 0\)versus\({{\rm{H}}_{\rm{a}}}{\rm{:\mu }} \ne {\rm{0}}\)

where\(\mu \) is the true average of the percentage differences.

The normal probability plot suggests that the t-test can be used here (perhaps the normality is not one hundred percent sure, however use the t-test to obtain the result).

The t statistic value can be computed using formula

\(t = \frac{{\bar x - {\Delta _0}}}{{s/\sqrt n }}\)

The Sample Mean\(\bar x\)of observations\({x_1},{x_2}, \ldots ,{x_n}\)is given by

\(\begin{array}{c}\bar x = \frac{{{x_1} + {x_2} + \ldots + {x_n}}}{n}\\ = \frac{1}{n}\sum\limits_{i = 1}^n {{x_i}} \end{array}\)

The sample mean\(\bar x\)is

\(\begin{array}{c}\bar x = \frac{1}{{10}} \times (17.78 + 45 + \ldots + 26.67)\\ = 27.29\% \end{array}\)

The Sample Variance\({s^2}\)is

\({s^2} = \frac{1}{{n - 1}} \times {S_{xx}}\)

where

\(\begin{array}{c}{S_{xx}} = \sum {{{\left( {{x_i} - \bar x} \right)}^2}} \\ = \sum {x_i^2} - \frac{1}{n} \times {\left( {\sum {{x_i}} } \right)^2}\end{array}\)

The Sample Standard Deviations is

\(s = \sqrt {{s^2}} = \sqrt {\frac{1}{{n - 1}} \times {S_{xx}}} .\)

The sample standard deviation is

\(\begin{array}{l} = \sqrt {\frac{1}{{10 - 1}}\left( {{{(17.78 - 27.29)}^2} + {{(45 - 27.29)}^2} + \ldots + {{(26.67 - 27.29)}^2}} \right)} \\\\ = 22.12\% \end{array}\)

Thus, the t statistic value is

\(\begin{array}{c}t = \frac{{27.29 - 0}}{{22.12/\sqrt {10} }}\\ = 3.9\end{array}\)

The degrees of freedom are \(n - 1 = 10 - 1 = 9.\) The P value for the two-sided alternative hypothesis is two times the area under the \({t_9}\) curve to the right of \(|t|\)

\(\begin{array}{c}P = 2 \times P(T > 3.9)\\ = 2 \times 0.002\\ = 0.004\end{array}\)

Where the value was computed using a software (you can use the table in the appendix of the book).

\(Since \)\(P = 0.004 < \alpha \)

Where \(\alpha \) is any reasonable significance level, reject null hypothesis

Since we reject the null hypothesis, the conclusion is that the true mean percentage difference between stated energy values and their measured energy value is not zero.