In: Statistics and Probability
1. You want to see if three different cafes yield different costs for a lunch meal for a week. You randomly select five measurements from trials on an automated driving machine for each cafe. At the 0.05 significance level, is there a difference among cafes in mean lunch meal cost? Test this claim showing all calculation steps clearly in the Ms World file (25P) and draw an ANOVA Table for the solution (10P). You have to use the same steps which we used in the lecture. You can find the steps from lecture slides or recorded video. Also, you have to take in a consider the HOMEWORK - PROJECT-PERFORMANCE EVALUATION FORM which is above. SEND IT IN MS WORD FORMAT NOT SCREEN SHOT
Cafe 1 |
Cafe 2 |
Cafe 3 |
25 |
23 |
26 |
23 |
28 |
24 |
25 |
25 |
26 |
28 |
23 |
22 |
24 |
16 |
28 |
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The hypothesis being tested is:
H0: µ1 = µ2 = µ3
Ha: Not all means are equal
Café 1 | Café 2 | Café 3 | |||
25 | 23 | 26 | |||
23 | 28 | 24 | |||
25 | 25 | 26 | |||
28 | 23 | 22 | |||
24 | 16 | 28 | |||
Mean | n | Std. Dev | |||
25.0 | 5 | 1.87 | Café 1 | ||
23.0 | 5 | 4.42 | Café 2 | ||
25.2 | 5 | 2.28 | Café 3 | ||
24.4 | 15 | 3.02 | Total | ||
ANOVA table | |||||
Source | SS | df | MS | F | p-value |
Treatment | 14.80 | 2 | 7.400 | 0.79 | .4773 |
Error | 112.80 | 12 | 9.400 | ||
Total | 127.60 | 14 |
The p-value is 0.4773.
Since the p-value (0.4773) is greater than the significance level (0.05), we fail to reject the null hypothesis.
Therefore, we cannot conclude that there is a difference among cafes in mean lunch meal costs.
Thank you! :)