In: Statistics and Probability
A project in Environment Studies requires you to study monthly home electrical usage. You randomly called 50 homes in Irvine asking how much each home pays on average for electricity during the fall. You were able to get only 20 responses with data displayed below. The population is assumed to be normally distributed. α = 0.01
68 85 60 72 42 55 40 64 67 60
52 72 110 70 68 56 35 75 50 84
Fill in the blank: We are ___________% confident that the interval from _________ to __________ actually contains _______________________________________________.
The sample size is n = 20 The provided sample data along with the data required to compute the sample mean and sample variance are shown in the table below:
X | X2 | |
68 | 4624 | |
85 | 7225 | |
60 | 3600 | |
72 | 5184 | |
42 | 1764 | |
55 | 3025 | |
40 | 1600 | |
64 | 4096 | |
67 | 4489 | |
60 | 3600 | |
52 | 2704 | |
72 | 5184 | |
110 | 12100 | |
70 | 4900 | |
68 | 4624 | |
56 | 3136 | |
35 | 1225 | |
75 | 5625 | |
50 | 2500 | |
84 | 7056 | |
Sum = | 1285 | 88261 |
The sample mean is computed as follows:
Also, the sample variance is
Therefore, the sample standard deviation s is
The number of degrees of freedom are df = 20 - 1 = 19 , and the significance level is α=0.01.
Based on the provided information, the critical t-value for α=0.01 and df = 19 degrees of freedom is t_c = 2.861
The 99% confidence for the population mean μ is computed using the following expression
Therefore, based on the information provided, the 99 % confidence for the population mean μ is
CI = (64.25 - 11.08, 64.25 + 11.08)
CI = (53.17, 75.33)
which completes the calculation.
We are 99% confident that the interval from 53.17 to 75.33 actually contains the amount home pays on average for electricity during the fall.