Question

In: Statistics and Probability

For each problem, perform the following steps. Assume that all variables are normally or approximately normally distributed.

 

For each problem, perform the following steps. Assume that all variables are normally or approximately normally distributed.

  1. State the hypothesis and identify the claim.

  1. Find the critical value(s).

  1. Compute the test value.

  1. Make the decision.

  1. Summarize the results.

  1. 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

Solutions

Expert Solution

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

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