A group of paramedics does not believe that the average number of calls in the m
ID: 3170925 • Letter: A
Question
A group of paramedics does not believe that the average number of calls in the morning, afternoon and night shifts are equal. They counted the number of calls over 7 days, and found the following:
Let Morning = group 1, Afternoon = group 2, and Night = group 3. Assume = 0.01.
(a) State the null and alternative hypothesis mathematically.
(b) State the null and alternative hypothesis in terms of the problem.
(c) Calculate the range for the p-value.
(d) Do you reject or fail to reject the null hypothesis?
(e) Interpret your conclusion in terms of the problem.
(f) Interpret a Type I error in terms of the problem.
We concluded there was a difference in the average number of calls of at least one of the shifts, when in reality there was not.
We concluded there was a difference in the average number of calls of all of the shifts, when in reality there was not.
We concluded there was not a difference in the average number of calls of the shifts, when in reality there was at least one difference.
Morning Afternoon Night Mean 2.57 3.71 4.29 Std. Dev 0.98 1.11 1.38 n 7 7 7 Ho 1Explanation / Answer
Answer
(a) State the null and alternative hypothesis mathematically.
In ANOVA, null hypothesis always focuses on equality of all treatment means and alternative hypothesis focuses on whether at least one of the treatment means significantly differs from other means.
Hence correct answer is –
H0: 1 = 2 = 3 vs HA: at least on i is different
(b) State the null and alternative hypothesis in terms of the problem.
On similar argument as in part a, we can see that correct answer is –
H0: The average number of calls for the time periods are equal vs HA: At least one average number of calls for the time periods is different.
(c) Calculate the range for the p-value.
Calculation of ANOVA table with P value is given below.
Source
Sum of squares
DF
Mean square
F0
P-value
Treatment
10.7203
2
5.3601
3.9250
0.0385
Error
24.5814
18
1.3656
Total
35.3017
20
It shows that, P value = 0.0385. Hence correct answer is –
0.02 < p-value < 0.05
(d) Do you reject or fail to reject the null hypothesis?
Since P value (0.0385) is greater than level of significance = .01, we fail to reject the null hypothesis.
Hence correct answer is –
Fail to reject
(e) Interpret your conclusion in terms of the problem.
Failing to reject the null hypothesis indicates that we cannot support alternative hypothesis and hence average number of calls between the shifts are equal.
Hence correct answer is –
We cannot support that the average number of calls between the shifts are different at 1% significance.
(f) Interpret a Type I error in terms of the problem.
Type I error means rejecting the null hypothesis when it is true. Here rejecting null hypothesis means concluding that there was a difference in the average number of calls of at least one of the shifts and null hypothesis being true means in reality all treatment means are equal.
Hence correct answer is –
We concluded there was a difference in the average number of calls of at least one of the shifts, when in reality there was not.
Source
Sum of squares
DF
Mean square
F0
P-value
Treatment
10.7203
2
5.3601
3.9250
0.0385
Error
24.5814
18
1.3656
Total
35.3017
20
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