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A small university is concerned with monitoring its electricity usage in its Student Center. Specifically, its officials want to know if the amount of electricity used differs by day of the week. They collected data for nearly a year, and the relevant summary statistics are provided. Note that electricity usage is measured in kilowatt hours. Day of Week n x bar s Sunday 45 86.48 34.89 Monday 45 109.29 27.37 Tuesday 45 110.96 28.64 Wednesday 44 115.03 31.68 Thursday 44 114.97 33.26 Friday 45 108.58 32.22 Saturday 45 87.07 38.56 Overall 313 104.56 34.25 Computer output from the analysis is provided: One-way ANOVA: Electricity versus Day of Week Source DF SS MS F P Day of Week 6 41646 6941 6.55 0.000 Error 306 324428 1060 Total 312 366073 What is the decision on the hypothesis test and why

Sagot :

Akiran007idn

Answer:

Since p-value > ∝, H0 is accepted

Step-by-step explanation:

Calculating gives the following results.

ANOVA Table

Source DF               Sum             Mean      F Statistic P-value

                                 of Square           Square

Groups     6         191.61                      31.935        0.020          0.999

(b/w groups)

Error

(within 14          21736.08                1552.577

groups)                                                                                                    

Total  20          21927.69 1096.385                                          

1. H0 hypothesis

Since p-value > ∝, H0 is accepted.

Hence the averages of all groups considered to be equal.

Or the difference between the averages of all groups is not big enough to reject H0.

2. P-value

P-value equals 0.999. This means that if we would reject H0, the chance of type1 error (rejecting a correct H0) would be too high i.e  (99.99%)

The bigger  p-value  supports H0.

3. The statistics

The test statistic F equals 0.020, which lies in the accepted range: [-∞ : 2.8477]

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