(a)
To calculate: The population mean for the confidence interval of 95%.
(a)
Answer to Problem 70E
Thepopulation mean is from0.0430% to 0.1178%.
Explanation of Solution
Given information:
Sample size
Confidence interval
Table for sample is:
Well | Bottom | Top |
1 | 0.430 | 0.415 |
2 | 0.226 | 0.238 |
3 | 0.567 | 0.390 |
4 | 0.531 | 0.410 |
5 | 0.707 | 0.605 |
6 | 0.716 | 0.609 |
7 | 0.651 | 0.632 |
8 | 0.589 | 0.523 |
9 | 0.469 | 0.411 |
10 | 0.723 | 0.612 |
Formula used:
Sample mean
Standard deviation
Degree of freedom
Margin of error
Calculation:
Find the mean, use the formula
Find the standard deviation, use the formula
Find the degree of freedom, use the formula
The confidence level is 95%.
Convert 95% into decimal.
Find the column.
From table B find critical value
Thus, critical value
Find the margin of error, use the formula
Now, find the population mean.
Hence, the required population mean is from 0.0430% to 0.1178%.
(b)
Explain whether or not the confidence interval gives convincing evidence of a difference in zinc concentrations at the top and bottom of wells in the region.
(b)
Answer to Problem 70E
The confidence interval is sufficient to give convincing evidence of a difference in zinc concentrations at the top and bottom of wells in the region.
Explanation of Solution
From 70(a)confidence interval contain the population mean from0.0430% to 0.1178%. Since the confidence interval lies above 0.
Hence, the confidence interval is sufficient to give convincing evidence of a difference in zinc concentrations at the top and bottom of wells in the region.
Chapter 8 Solutions
The Practice of Statistics for AP - 4th Edition
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Statistics for Business and Economics (13th Edition)
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