Save Homework: HW 8 30f 4 (4 complete) Hw score: 75%, 3 of 4 Score: 0.17 of 1 pt

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Save Homework: HW 8 30f 4 (4 complete) Hw score: 75%, 3 of 4 Score: 0.17 of 1 pt i Question Help 163.49-P To see how much difference time of day made on theEarly Evening Late night speed at which he could download files, a college (7 am) (5 p.m.) (12 a.m) sophomore placed a file on a remote server, then proceeded to download it at three different time periods of the day. He downloaded the file 18 times in all, 6 tires at each time of day, and recorded the time in seconds that the download tool. At the 5% significance level, do the data provide sufficient evidence to conclude that a difference exists in mean 84 sec 205 sec 165 sec download speed? 65 sec 208 sec 210 sec 142 sec 311 sec 176 sec 94 sec 256 sec | 176 sec 203 sec 325 sec 227 sec 110 sec 228 sec 212 sec Let 1 2 and 3 be the population means times for 7 a m 5 pm and 12 a , respectively what are the correct hypotheses for a one-way ANOVA test? Ha: All the means are equal Ha: Not all the means are equal. What is the F-statistic? Fe(Round to two decimal places as needed ) Enter your answer in the answer box and then click Check Answer parts remaining Clear All Check Answer

Explanation / Answer

Solution:

Here, we have to use one way analysis of variance or ANOVA F test for checking whether the given three population means are same or not.

H0: The average downloading time is same at the early morning, evening, and late night.

Ha: The average downloading time is not same at the early morning, evening, and late night.

We are given level of significance as 5% or = 0.05.

From the given data, we have descriptive statistics as below:

Groups

Count

Sum

Average

Variance

Early

6

698

116.3333333

2477.8667

Evening

6

1533

255.5

2694.7000

Late night

6

1075

179.1666667

1368.5667

By using the given data and above descriptive statistics, the ANOVA table is given as below:

Total sample size = 6+6+6 = 18

Total df = N – 1 = 18 – 1 = 17

Total number of groups = 3

df for between groups = 3 – 1 = 2

df for within groups = total df – between df = 17 – 2 = 15

MS = SS/df

F = MSB/MSW

F = 29142.1667/2180.3778 = 13.3657

F = 13.37

ANOVA

Source of Variation

SS

df

MS

F

P-value

Between Groups

58284.3333

2

29142.1667

13.3657

0.0005

Within Groups

32705.6667

15

2180.3778

Total

90990.0000

17

(Values for SS and P-value are calculated by using excel.)

Here, p-value < = 0.05.

So, we reject the null hypothesis that the average downloading time is same at the early morning, evening, and late night.

There is sufficient evidence to conclude that the average downloading time is not same at the early morning, evening, and late night.

Groups

Count

Sum

Average

Variance

Early

6

698

116.3333333

2477.8667

Evening

6

1533

255.5

2694.7000

Late night

6

1075

179.1666667

1368.5667

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