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The attractiveness ratings given by participants in a speed dating session are provided.

It is claimed that there is a difference between male attractiveness ratings and female attractiveness ratings.

Null Hypothesis: The mean of the differences between male and female attractiveness ratings is equal to 0.

\({H_0}:{\mu _d} = 0\)

Alternative Hypothesis: The mean of the differences between male and female attractiveness ratings is not equal to 0.

\({H_1}:{\mu _d} \ne 0\)

The differences between male and female attractiveness ratings are tabulated below:

\(\begin{array}{c}\overline d = \frac{{\sum\limits_{i = 1}^n {{d_i}} }}{n}\\ = \frac{{\left( { - 2} \right) + 0 + ..... + 2}}{{12}}\\ = - 0.208\end{array}\)

The mean of the differences is computed below:

The standard deviation of the differences is computed below:

\(\begin{array}{c}{s_d} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({d_i} - \overline d )}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {\left( { - 2} \right) - \left( { - 0.208} \right)} \right)}^2} + {{\left( {0 - \left( { - 0.208} \right)} \right)}^2} + ..... + {{\left( {2 - \left( { - 0.208} \right)} \right)}^2}}}{{12 - 1}}} \\ = 2.069\end{array}\)

Thus, the mean and standard deviation of the differences are -0.208 and 2.069, respectively.

The assumption from the null hypothesis states that \({\mu _d} = 0\).

Then, the test statistic value is computed below:

\(\begin{array}{c}t = \frac{{\overline d - {\mu _d}}}{{\frac{{{s_d}}}{{\sqrt n }}}}\\ = \frac{{ - 0.208 - 0}}{{\frac{{2.169}}{{\sqrt {12} }}}}\\ = - 0.333\end{array}\)

Thus, the value of t is equal to -0.333.

The value of the degrees of freedom is computed below:

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

Using the t-distribution table, the critical values of t for 11 degrees of freedom and at a 0.05 level of significance are -2.201 and 2.201.

The two-tailed p-value of the z-score equal to -0.333 is equal to 0.7454.

Since the test statistic value lies within the two critical values and the p-value is greater than 0.05, the null hypothesis is failed to reject.

There is insufficient evidence to support the claim that there is a difference between female attractiveness ratings and male attractiveness ratings.