Speed Dating

Listed below are attribute ratings of males by females who participated in speed dating events (from Data Set 18 “Speed Dating” in Appendix B). Use a 0.05 significance level to test the claim that females in the different age brackets give attribute ratings with the same mean. Does age appear to be a factor in the female attribute ratings?

Age 20-22	38	42	30.0	39	47	43	33	31	32	28
Age 23-26	39	31	36.0	35	41	45	36	23	36	20
Age 27-29	36	42	35.5	27	37	34	22	47	36	32

The data set all have the same sample size. The given sample size \[n = 10\]\(\left( {{n_1} = {n_2} = {n_3} = 10} \right)\).

Number of samples (k) \( = 3\)

Significance level,\(\alpha \)\( = 0.05\)

To test the claim that females in different age brackets give attribute ratings with the same mean, the null hypothesis is formulated as,

\({H_0}:{\mu _1} = {\mu _2} = {\mu _3}\)

The corresponding alternative hypothesis is

\[{H_1}:\]At least one of the means is different from the others.

Let,

\[{\bar x_1},{\bar x_2}{\rm{ and}}\;{\bar x_3}\]denotes the sample means and

\({n_1},{n_2}{\rm{ and }}{n_3}\)denotes the sample sizes.

Then,

\(\begin{aligned}{c}{{\bar x}_1} = \frac{1}{{{n_1}}}\sum\limits_{i = 1}^n {{x_i}} \\ = \frac{{38 + 42 + 30 + 39 + 47 + 43 + 33 + 31 + 32 + 28}}{{10}}\\ = \frac{{363}}{{10}}\\ = 36.3\end{aligned}\)

\(\begin{aligned}{c}{{\bar x}_2} = \frac{1}{{{n_2}}}\sum\limits_{i = 1}^n {{x_i}} \\ = \frac{{39 + 31 + 36 + 35 + 41 + 45 + 36 + 23 + 36 + 20}}{{10}}\\ = \frac{{342}}{{10}}\\ = 34.2\end{aligned}\)

\(\begin{aligned}{c}{{\bar x}_3} = \frac{1}{{{n_3}}}\sum\limits_{i = 1}^n {{x_i}} \\ = \frac{{36 + 42 + 35.5 + 27 + 37 + 34 + 22 + 47 + 36 + 32}}{{10}}\\ = \frac{{348.5}}{{10}}\\ = 34.85\end{aligned}\)

Thus, the sample means are 36.3, 34.2, and 34.85.

Let \(s_1^2,s_2^2{\rm{ and }}s_3^2\)denotes the sample variances. Then,

\(\begin{aligned}{c}s_1^2 = \frac{1}{{{n_1} - 1}}{\left( {\sum\limits_{i = 1}^n {{x_i}} - {{\bar x}_1}} \right)^2}\\ = \frac{1}{9}\left[ \begin{aligned}{l}{\left( {38 - 36.3} \right)^2} + {\left( {42 - 36.3} \right)^2} + {\left( {30 - 36.3} \right)^2} + {\left( {39 - 36.3} \right)^2} + {\left( {47 - 36.3} \right)^2} + {\left( {43 - 36.3} \right)^2}\\ + {\left( {33 - 36.3} \right)^2} + {\left( {31 - 36.3} \right)^2} + {\left( {32 - 36.3} \right)^2} + {\left( {28 - 36.3} \right)^2}\end{aligned} \right]\\ = 40.9\end{aligned}\)\(\)

\(\begin{aligned}{c}s_2^2 = \frac{1}{{{n_2} - 1}}{\left( {\sum\limits_{i = 1}^n {{x_i}} - {{\bar x}_2}} \right)^2}\\ = \frac{1}{9}\left[ \begin{aligned}{l}{\left( {39 - 34.2} \right)^2} + {\left( {31 - 34.2} \right)^2} + {\left( {36 - 34.2} \right)^2} + {\left( {35 - 34.2} \right)^2} + {\left( {41 - 34.2} \right)^2} + {\left( {45 - 34.2} \right)^2}\\ + {\left( {36 - 34.2} \right)^2} + {\left( {23 - 34.2} \right)^2} + {\left( {36 - 34.2} \right)^2} + {\left( {20 - 34.2} \right)^2}\end{aligned} \right]\\ = 59.28\end{aligned}\)

\(\begin{aligned}{c}s_3^2 = \frac{1}{{{n_3} - 1}}{\left( {\sum\limits_{i = 1}^n {{x_i}} - {{\bar x}_3}} \right)^2}\\ = \frac{1}{9}\left[ \begin{aligned}{l}{\left( {36 - 34.85} \right)^2} + {\left( {42 - 34.85} \right)^2} + {\left( {35.5 - 34.85} \right)^2} + {\left( {27 - 34.85} \right)^2} + {\left( {37 - 34.85} \right)^2} + {\left( {34 - 34.85} \right)^2}\\ + {\left( {22 - 34.85} \right)^2} + {\left( {47 - 34.85} \right)^2} + {\left( {36 - 34.85} \right)^2} + {\left( {32 - 34.85} \right)^2}\end{aligned} \right]\\ = 49.11\end{aligned}\)

Thus, the sample variances are 40.9, 59.28, and 49.11.

Let,

\(\bar \bar x\)denotes the mean of sample means.

\(s_{\bar X}^2\)denotes the variance of sample means.

\(\begin{aligned}{c}\bar \bar x = \frac{1}{k}\sum\limits_{i = 1}^n {{{\bar x}_i}} \\ = \frac{{36.3 + 34.2 + 34.85}}{3}\\ = 35.11\end{aligned}\)

\(\begin{aligned}{c}s_{\bar X}^2 = \frac{1}{{k - 1}}\sum\limits_{i = 1}^n {{{({{\bar x}_i} - \bar \bar x)}^2}} \\ = \frac{1}{2}\left[ {{{\left( {36.3 - 35.11} \right)}^2} + {{\left( {34.2 - 35.11} \right)}^2} + {{\left( {34.85 - 35.11} \right)}^2}} \right]\\ = 1.16\end{aligned}\)

Thus, the mean of the sample mean is 35.11, and the variance of the sample mean is 1.16.

Let \(ns_{\bar X}^2\)denotes the variance between sample means.

Then,

\(\begin{aligned}{c}ns_{\bar X}^2 = 10\left( {1.16} \right)\\ = 11.6\end{aligned}\)

Thus, the variance between samples is 11.6.

Let \(s_p^2\)denotes the variance within samples, then

\(\begin{aligned}{c}s_p^2 = \frac{1}{n}\sum\limits_{i = 1}^n {s_i^2} \\ = \frac{1}{3}\left[ {40.9 + 59.28 + 49.11} \right]\\ = 49.76\end{aligned}\)

Thus, the variance within samples is 49.76.

The F test statistic is

\(\begin{aligned}{c}F = \frac{{{\rm{Variance between samples}}}}{{{\rm{variance within samples}}}}\\ = \frac{{ns_{\bar X}^2}}{{s_p^2}}\\ = \frac{{11.6}}{{49.76}}\\ = 0.2331\end{aligned}\)

Thus, the value of F is 0.2331.

Let, \(k\)is the number of samples, and n is the sample size. Then,

Numerator degrees of freedom = \(k - 1\)

Denominator degrees of freedom= \(k\left( {n - 1} \right)\)

Here,

\(k - 1 = 2\)and \(k\left( {n - 1} \right) = 27\).

With \(\alpha = 0.05\), the critical value of F is

\({F_{\left( {2,27} \right)}} = 3.{\rm{3541}}\)\(\left( {{\rm{from F - distribution table}}} \right)\)

Here, \(F < {F_{\left( {2,27} \right)}}\), the null hypothesis fails to reject. So, there is enough t evidence to support the claim that females in different age brackets give attributes with the same means. Thus, it can be said that age does not appear to be a factor in the female attribute ratings.