In: Statistics and Probability
For each problem, perform the following steps. Assume that all variables are normally or approximately normally distributed.
State the hypothesis and identify the claim.
Find the critical value(s).
Compute the test value.
Make the decision.
Summarize the results.
The average temperatures for a 25-day period for Birmingham, Alabama, and Chicago, Illinois, are shown. Based on the samples, at α = 0.10, can it be concluded that it is warmer in Birmingham? [4]
Birmingham |
Chicago |
||||||||
78 |
82 |
68 |
67 |
68 |
70 |
74 |
73 |
60 |
77 |
75 |
73 |
75 |
64 |
68 |
71 |
72 |
71 |
74 |
76 |
62 |
73 |
77 |
78 |
79 |
71 |
80 |
65 |
70 |
83 |
74 |
72 |
73 |
78 |
68 |
67 |
76 |
75 |
62 |
65 |
73 |
79 |
82 |
71 |
66 |
66 |
65 |
77 |
66 |
64 |
Let be the the average temperature in Birmingham, Alabama.
Let be the the average.in Chicago, Illinois.
From the data, the following was calculated.
For: = 72.92, s1 = 5.5, n1 = 25
For: = 72.8, s2 = 5.82, n2 = 25
Since s1/s2 = 5.5 / 5.82 = 0.95 (it lies between 0.5 and 2) we used the pooled variance.
The degrees of freedom used is n1 + n2 - 2 = 25 + 25 - 2 = 48 (since pooled variance is used)
_________________________
The Hypothesis:
H0: : The temperatures in the 2 cities are the same.
Ha: : Birmingham is warmer than Illinois. (Claim)
This is a Right tailed test.
_______________________
The Critical Value: The critical value (Right tail) at = 0.10, df = 48, tcritical = + 1.2994
Therefore Reject H0, If t observed is > 1.2994
_____________________
The Test Statistic:
______________________________________
The Decision: Since t observed (1.32) is > t critical (1.299), We Reject H0.
________________________________________
The p Value: The p value (Right Tail) for t = 1.32, df = 48, is; p value = 0.0959
_______________________________________________________
The Conclusion: There is sufficient evidence at the 90% significance level to conclude that Birmingham, Alabama is warmer than Chicago, Illinois
_______________________________________________________
Calculation for the mean and standard deviation:
Mean = Sum of observation / Total Observations
Standard deviation = SQRT(Variance)
Variance = Sum Of Squares (SS) / n - 1, where SS = SUM(X - Mean)2.
Birmingham | Chicago | |||||||
# | X | Mean | (X-Mean)2 | # | X | Mean | (X-Mean)2 | |
1 | 78 | 72.92 | 25.8064 | 1 | 70 | 70.8 | 0.64 | |
2 | 75 | 72.92 | 4.3264 | 2 | 71 | 70.8 | 0.04 | |
3 | 62 | 72.92 | 119.2464 | 3 | 71 | 70.8 | 0.04 | |
4 | 74 | 72.92 | 1.1664 | 4 | 67 | 70.8 | 14.44 | |
5 | 73 | 72.92 | 0.0064 | 5 | 66 | 70.8 | 23.04 | |
6 | 82 | 72.92 | 82.4464 | 6 | 74 | 70.8 | 10.24 | |
7 | 73 | 72.92 | 0.0064 | 7 | 72 | 70.8 | 1.44 | |
8 | 73 | 72.92 | 0.0064 | 8 | 80 | 70.8 | 84.64 | |
9 | 72 | 72.92 | 0.8464 | 9 | 76 | 70.8 | 27.04 | |
10 | 79 | 72.92 | 36.9664 | 10 | 65 | 70.8 | 33.64 | |
11 | 68 | 72.92 | 24.2064 | 11 | 73 | 70.8 | 4.84 | |
12 | 75 | 72.92 | 4.3264 | 12 | 71 | 70.8 | 0.04 | |
13 | 77 | 72.92 | 16.6464 | 13 | 65 | 70.8 | 33.64 | |
14 | 73 | 72.92 | 0.0064 | 14 | 75 | 70.8 | 17.64 | |
15 | 82 | 72.92 | 82.4464 | 15 | 77 | 70.8 | 38.44 | |
16 | 67 | 72.92 | 35.0464 | 16 | 60 | 70.8 | 116.64 | |
17 | 64 | 72.92 | 79.5664 | 17 | 74 | 70.8 | 10.24 | |
18 | 78 | 72.92 | 25.8064 | 18 | 70 | 70.8 | 0.64 | |
19 | 78 | 72.92 | 25.8064 | 19 | 62 | 70.8 | 77.44 | |
20 | 71 | 72.92 | 3.6864 | 20 | 66 | 70.8 | 23.04 | |
21 | 68 | 72.92 | 24.2064 | 21 | 77 | 70.8 | 38.44 | |
22 | 68 | 72.92 | 24.2064 | 22 | 76 | 70.8 | 27.04 | |
23 | 79 | 72.92 | 36.9664 | 23 | 83 | 70.8 | 148.84 | |
24 | 68 | 72.92 | 24.2064 | 24 | 65 | 70.8 | 33.64 | |
25 | 66 | 72.92 | 47.8864 | 25 | 64 | 70.8 | 46.24 |
Birmingham | Chicago | |
n | 25 | 25 |
Sum | 1823 | 1770 |
Mean | 72.92 | 70.80 |
SS | 725.84 | 812 |
Variance | 30.24 | 33.83 |
SD | 5.50 | 5.82 |