Concept explainers
a.
Find the population
a.
Answer to Problem 19E
The population mean
Explanation of Solution
Calculation:
The given data shows that the summer temperature for five days in July.
The formula to find the population mean is,
Thus, the population mean
The formula to find the population standard deviation is,
Here, x represents the population values,µ represents the population mean and N represents the population size.
Substitute µ as75.4 and N as5.
That is,
Thus, the standard deviation
b.
List all samples of size 2 drawn with replacement.
b.
Explanation of Solution
Calculation:
The two values are randomly selected with replacement from the population {69, 75, 79, 83, 71}.
Thedifferent possible samples of size 2 are {(69, 69), (69,75), (69, 79), (69, 83), (69, 71), (75, 69), (75, 75), (75, 79), (75, 83), (75, 71), (79, 69), (79, 75), (79, 79), (79, 83), (79, 71), (83, 69), (83, 75), (83, 79), (83, 83), (83, 71), (71, 69), (71, 75), (71, 79), (71, 83), (71, 71)}.
Thus, there are 25 different samples of size 2.
c.
Compute the sample mean
c.
Answer to Problem 19E
The sample mean
Sample | Sample mean |
69, 69 | 69 |
69, 75 | 72 |
69, 79 | 74 |
69, 83 | 76 |
69, 71 | 70 |
75, 69 | 72 |
75, 75 | 75 |
75, 79 | 77 |
75, 83 | 79 |
75, 71 | 73 |
79, 69 | 74 |
79, 75 | 77 |
79, 79 | 79 |
79, 83 | 81 |
79, 71 | 75 |
83, 69 | 76 |
83, 75 | 79 |
83, 79 | 81 |
83, 83 | 83 |
83, 71 | 77 |
71, 69 | 70 |
71, 75 | 73 |
71, 79 | 75 |
71, 83 | 77 |
71, 71 | 71 |
Total | 1,885 |
The mean
Explanation of Solution
Calculation:
The formula for finding sample mean is
The sample mean for the sample (69, 69):
The formula to find the sample variance is,
Where, x represents the sample values,
The sample variance for the sample (69, 69):
Substitute
That is,
Similarly, the sample mean and sample standard deviation for the remaining samples of size 2 is obtained as given in below Table.
Sample | Sample mean | Sample variance |
69, 69 | 69 | 0 |
69, 75 | 72 | 18 |
69, 79 | 74 | 50 |
69, 83 | 76 | 98 |
69, 71 | 70 | 2 |
75, 69 | 72 | 18 |
75, 75 | 75 | 0 |
75, 79 | 77 | 8 |
75, 83 | 79 | 32 |
75, 71 | 73 | 8 |
79, 69 | 74 | 50 |
79, 75 | 77 | 8 |
79, 79 | 79 | 0 |
79, 83 | 81 | 8 |
79, 71 | 75 | 32 |
83, 69 | 76 | 98 |
83, 75 | 79 | 32 |
83, 79 | 81 | 8 |
83, 83 | 83 | 0 |
83, 71 | 77 | 72 |
71, 69 | 70 | 2 |
71, 75 | 73 | 8 |
71, 79 | 75 | 32 |
71, 83 | 77 | 72 |
71, 71 | 71 | 0 |
Total | 1,885 | 656 |
The mean
Thus, the mean
The standard deviation
Thus, the standard deviation
d.
Verify that
d.
Explanation of Solution
Calculation:
The two values are randomly selected with replacement. Therefore, n is 2.
From part (a), the population mean
Substitute 75.4 for
Substitute 5.1225 for
From part (c), the mean
Thus, the mean and standard deviation both are equal.
Hence
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Chapter 6 Solutions
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