Question

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Three different gyms survey some of their gym members. The three different gyms have differnt types of work outs, Gym A offers a HIIT workout, Gym B offers a Bootcamp style workout, and Gym C offers a Cardio workout The Gyms ask their members to disclose how long they worked out at the gym on a given Monday. The minutes that they worked out in the gym, on that Monday, were recorded. Gym B Gym A Gym C 43 64 46 28 27 54 41 30 41 36 44 64 46 61 56 46 65 36 37 20 55 18 19 Assume that all distributions are normal, the three population standard deviations are all the same, and the data was collected independently and randomly. Use a level of significance of a = 0.1. Ho: Mi = H2 = 43 H: At least two of the means differ from each other. 1. For this study, we should use Select an answer 2. The test statistic for this data (Please round your answer to 3 decimal places.) 3. The p-value for this sample - (Please round your answer to 4 decimal places.) 4. The p-value is Select an answer va 5. Base on this, we should Select an answer hypothesis 6. As such, the final conclusion is that... There is insufficient evidence to support the claim that Gym Workout Type is a factor in minutes gym members worked out in the gym. O There is sufficient evidence to support the claim that Gym Workout Type is a factor in minutes gym members worked out in the gym.

Answer 1

The following table is obtained: Group 1 Group 2 Group 3 43 64 46 28 27 54 41 30 41 36 44 64 46 61 56 46 65 36 37 20 55 18 19 Sum = 314 311 352 Average = 34.889 44.429 50.286 11896 16007 18266 St. Dev. = 10.845 19.104 9.708 SS = 940.888888888 89 2189.71428571 43 565.428571428 57 n = 9 7 7 The total sample size is N = 23. Therefore, the total degrees of freedom are: dftotal = 23 – 1= 22 Also, the between-groups degrees of freedom are dfbetween = 3 – 1 = 2, and the within-groups degrees of freedom are: dfwithin = dftotal – d fbetween = 22 – 2 = 20 First, we need to compute the total sum of values and the grand mean. The following is obtained Xij = 314 + 311 + 352 = 977 Also, the sum of squared values is = 11896 + 16007 + 18266 = 46169 Based on the above calculations, the total sum of squares is computed as follows S Stotal = x; = 46169 9772 23 = 4667.739 The within sum of squares is computed as shown in the calculation below: SSwithin = sswithingroups 940.88888888889 + 2189.7142857143 + 565.42857142857 = 3696.032

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: 971.707 M Sbetween SSbetween dfbetween = 485.854 2 SS within MS within 3696.032 20 = 184.802 d fwithin Now that sum of squares are computed, we can proceed with computing the mean sum of squares: M Sbetween S Sbetween d fbetween 971.707 2 485.854 MS within S Swithin dfwithin 3696.032 20 184.802 Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows: F= M Sbetween M Swithin 485.854 184.802 = 2.629 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: M1 = H2 = M3 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 a = 0.1, and the degrees of freedom are dfi = 2 and df2 = 2, therefore, the rejection region for this F-test is R= = {F: F > Fc = 2.589}. (3) Test Statistics F= M Sbetween M Swithin 485.854 184.802 2.629 (4) Decision about the null hypothesis Since it is observed that F = 2.629 > Fc = 2.589, it is then concluded that the null hypothesis is rejected. = 0.0969 Using the P-value approach: The p-value is p 0.0969, and since p 0.1, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that not all 3 population means are equal, at the a = 0.1 significance level.

1. For this study, we should use ANOVA.

2. The t- statistic for this data = 2.629.

3. The P-value for this sample = 0.0969.

4. The P-value is less than $\large \alpha$ .

5. Base on this, we should reject the null hypothesis.

6. As such the final conclusion is that...

There is insufficient evidence to support the claim that Gym Workout Type is a factor in minutes gym members worked out in the gym.

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