Question

In Exercises 5–16, use analysis of variance for the indicated test.

Triathlon Times Jeff Parent is a statistics instructor who participates in triathlons. Listed below are times (in minutes and seconds) he recorded while riding a bicycle for five stages through each mile of a 3-mile loop. Use a 0.05 significance level to test the claim that it takes the same time to ride each of the miles. Does one of the miles appear to have a hill?

Mile 1	3:15	3:24	3:23	3:22	3:21
Mile 2	3:19	3:22	3:21	3:17	3:19
Mile 3	3:34	3:31	3:29	3:31	3:29

Short answer

No, the 3 miles do not appear to have the same duration of ride.

The third mile appears to have a hill.

Step-by-step solution

Given information

Data are given on the riding duration of five stages in three different miles of JeffParent.

State the hypotheses

Let\({\mu _1},{\mu _2},{\mu _3}\)be the actual population mean for the three miles.

The null hypothesis to test the difference in the riding duration of the 3 samples is as follows:

\(\begin{aligned}{l}{H_0}:{\mu _1} = {\mu _2} = {\mu _3}\\{H_a}:{\rm{At}}\;{\rm{least}}\;{\rm{one}}\;{\rm{mean}}\;{\rm{is}}\;{\rm{different}}\;{\rm{from}}\;{\rm{others}}\end{aligned}\)

State the decision rule

If the computed F-statistic is greater than the critical value, the null hypothesis is rejected at a 5% level of significance.

If the computed F-statistic is less than the critical value, the null hypothesis is failed to be rejected at a 5% level of significance.

Compute the variance between the groups a

Convert the samples into seconds as shown below:

3:15 is converted as\[3 \times 60 + 15 = 195\]. Similarly, the values in terms of seconds are shown below,

Mile 1	195	204	203	202	201
Mile 2	199	202	201	197	199
Mile 3	214	211	209	211	209

Let n denote the sample sizes.

As the 3 samples are of equal size,\(n = 5\).

Let\({\bar x_i}\)denote the sample means.

The 3 sample means are computed below:

\[\begin{aligned}{c}{{\bar x}_1} = \frac{{195 + 204 + 203 + 202 + 201}}{5}\\ = 201\\{{\bar x}_2} = \frac{{199 + 202 + 201 + 197 + 199}}{5}\\ = 199.60\\{{\bar x}_3} = \frac{{214 + 211 + 209 + 211 + 209}}{5}\\ = 210.80\\\end{aligned}\]

Let\(s_{\bar x}^2\)denote the variance of the sample means:

The mean of the 3 sample means is equal to:

\(\begin{aligned}{c}\bar \bar x = \frac{{201 + 199.60 + 210.80}}{3}\\ = 203.80\end{aligned}\)

The variance of the sample means is computed below:

\(\begin{aligned}{c}s_{\bar x}^2 = \frac{{\sum\limits_{i = 1}^3 {{{({{\bar x}_i} - \bar \bar x)}^2}} }}{{3 - 1}}\\ = \frac{{{{\left( {201 - 203.80} \right)}^2} + {{\left( {199.60 - 203.80} \right)}^2} + {{\left( {210.80 - 203.80} \right)}^2}}}{{3 - 1}}\\ = 37.24\end{aligned}\)

The variance between sample means is equal to:

\(\begin{aligned}{c}ns_{\bar x}^2 = 5\left( {37.24} \right)\\ = 186.20\end{aligned}\)

Therefore, the variance between sample means is equal to 186.20.

Compute the variance within the groups

Now, the 3 sample variances denoted by\({s_i}^2\)are computed below:

\(\begin{aligned}{c}{s_1}^2 = \frac{{\sum\limits_{i = 1}^n {{{({x_{1i}} - {{\bar x}_1})}^2}} }}{{n - 1}}\\ = \frac{{{{\left( {195 - 201} \right)}^2} + {{\left( {204 - 201} \right)}^2} + {{\left( {203 - 201} \right)}^2} + {{\left( {202 - 201} \right)}^2} + {{\left( {201 - 201} \right)}^2}}}{{5 - 1}}\\ = 12.50\end{aligned}\)

\(\begin{aligned}{c}{s_2}^2 = \frac{{\sum\limits_{i = 1}^n {{{({x_{2i}} - {{\bar x}_2})}^2}} }}{{n - 1}}\\ = \frac{{{{\left( {199 - 199.60} \right)}^2} + {{\left( {202 - 199.60} \right)}^2} + {{\left( {201 - 199.60} \right)}^2} + {{\left( {197 - 199.60} \right)}^2} + {{\left( {199 - 199.60} \right)}^2}}}{{5 - 1}}\\ = 3.80\end{aligned}\)

\(\begin{aligned}{c}{s_3}^2 = \frac{{\sum\limits_{i = 1}^n {{{({x_{3i}} - {{\bar x}_3})}^2}} }}{{n - 1}}\\ = \frac{{{{\left( {214 - 210.80} \right)}^2} + {{\left( {211 - 210.80} \right)}^2} + {{\left( {209 - 210.80} \right)}^2} + {{\left( {211 - 210.80} \right)}^2} + {{\left( {209 - 210.80} \right)}^2}}}{{5 - 1}}\\ = 4.20\end{aligned}\)

The variance within samples\(\left( {{s_p}^2} \right)\)is equal to:

\(\begin{aligned}{c}{s_p}^2 = \frac{{{s_1}^2 + {s_2}^2 + {s_3}^2}}{3}\\ = \frac{{12.50 + 3.80 + 4.20}}{3}\\ = 6.83\end{aligned}\)

Therefore, the variance within samples is equal to 6.83.

Compute the test statistic

Now, the F-statistic is computed as shown below:

\(\begin{aligned}{c}F = \frac{{{\rm{variance}}\;{\rm{between}}\;{\rm{samples}}}}{{{\rm{variance}}\;{\rm{within}}\;{\rm{samples}}}}\\ = \frac{{ns_{\bar x}^2}}{{{s_p}^2}}\\ = \frac{{186.20}}{{6.83}}\\ = 27.249\end{aligned}\)

Therefore, the value of the F-statistic is equal to 27.249.

Let k be the number of samples.

Thus, the value of k is 3.

Thus, the degrees of freedom are calculated as shown below:

\(\begin{aligned}{c}df = \left( {k - 1,k\left( {n - 1} \right)} \right)\\ = \left( {3 - 1,3\left( {5 - 1} \right)} \right)\\ = \left( {2,12} \right)\end{aligned}\)

Make a decision

The critical value of F at a 0.05 level of significance with (2,12) degrees of freedom from the F-distribution table is equal to 3.8853.

It can be observed that:

\(\begin{aligned}{c}F > {F_{crit}}\\\left( {27.249} \right) > \left( {3.885} \right)\end{aligned}\)

Since the value of the F-statistic is greater than the critical value, the null hypothesis is rejected at a 5% level of significance.

Therefore, there is enough evidence to conclude that there is a significant difference in the duration of the ride for Jeff Parent.

In other words, the 3 miles do not appear to have the same duration of the ride. At least one mean measure is different from others.

Also, it can be observed from the mean of the third mile that it takes a longer time and has a hill.