Question

An employee of a small software company in Minneapolis bikes to work during the summer months. He can travel to work using one of three routes and wonders whether the average commute times (in minutes) differ between the three routes. He obtains the following data after traveling each route for one week.

Route 1	32	34	37	32	26
Route 2	23	22	30	20	24
Route 3	20	27	25	25	27

Click here for the Excel Data File

a-1. Construct an ANOVA table. (Round intermediate calculations to at least 4 decimal places. Round "SS", "MS", "p-value" to 4 decimal places and "F" to 3 decimal places.)

a-2. At the 1% significance level, do the average commute times differ between the three routes. Assume that commute times are normally distributed.

• Yes   since the p-value is less than significance level.
• No   since the p-value is less than significance level.
• No   since the p-value is not less than significance level.
• Yes   since the p-value is not less than significance level.

b. Use Tukey’s HSD method at the 1% significance level to determine which routes' average times differ. (You may find it useful to reference the q table). (If the exact value for n_T − c is not found in the table, use the average of corresponding upper & lower studentized range values. Negative values should be indicated by a minus sign. Round your answers to 2 decimal places.)

Answer 1

The total sample size is N = 15N=15. Therefore, the total degrees of freedom are:

Also, the between-groups degrees of freedom are , and the within-groups degrees of freedom are:

First, we need to compute the total sum of values and the grand mean. The following is obtained

Also, the sum of squared values is

Based on the above calculations, the total sum of squares is computed as follows

The within sum of squares is computed as shown in the calculation below:

The between sum of squares is computed directly as shown in the calculation below:

The between sum of squares is computed directly as shown in the calculation below:

Now that sum of squares are computed, we can proceed with computing the mean sum of squares:

Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows:

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ₁ = μ₂ = μ₃

Ha: Not all means are equal

The above hypotheses will be tested using an F-ratio for a One-Way ANOVA.

(2) Rejection Region

Based on the information provided, the significance level is \alpha = 0.01α=0.01, and the degrees of freedom are df_1 = 2df1​=2 and df_2 = 2df2​=2, therefore, the rejection region for this F-test is

(3) Test Statistics

(4) Decision about the null hypothesis

Since it is observed that , it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.0148, and since , it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that not all 3 population means are equal, at the α=0.01 significance level.

So,

No, since p-value is not less than significance level.

b. since, means are not different at first place, it rules out the possibility of comparing each one of them seperately or we can say that none of the 3 Routes are significantly different.

Please rate my answer and comment for doubt.