Question

In Exercises 5–16, use the listed paired sample data, and assume that the samples are simple random samples and that the differences have a distribution that is approximately normal.

Speed Dating: Attractiveness Listed below are “attractiveness” ratings made by participantsin a speed dating session. Each attribute rating is the sum of the ratings of five attributes (sincerity, intelligence, fun, ambition, shared interests). The listed ratings are from Data Set 18 “Speed Dating.” Use a 0.05 significance level to test the claim that there is a difference between female attractiveness ratings and male attractiveness ratings.

Rating of Male by Female	4.0	8.0	7.0	7.0	6.0	8.0	6.0	4.0	2.0	5.0	9.5	7.0
Rating of Female by Male	6.0	8.0	7.0	9.0	5.0	7.0	5.0	4.0	6.0	8.0	6.0	5.0

Short answer

There is not sufficient evidence to support the claim that there is a significant difference between female attractiveness ratings and male attractiveness ratings.

Step-by-step solution

Given information

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.

Defining the hypotheses

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.

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

Compute the mean and standard deviation of the differences

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

Rating of Male by Female	4.0	8.0	7.0	7.0	6.0	8.0	6.0	4.0	2.0	5.0	9.5	7.0
Rating of Female by Male	6.0	8.0	7.0	9.0	5.0	7.0	5.0	4.0	6.0	8.0	6.0	5
Difference	-2	0	0	-2	1	1	1	0	-4	-3	3.5	2

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

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.

Compute the test statistic

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.

State the critical values and the p-value

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.

Conclusion

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.