Mr. Ahmed has 9 water heaters for his apartment buildings. each heater weighs 32 kilograms. How many grams do all 9 water heater
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Answer:
a) The 95% CI for the true average porosity is (4.51, 5.19).
b) The 98% CI for true average porosity is (4.11, 5.01)
c) A sample size of 15 is needed.
d) A sample size of 101 is needed.
Step-by-step explanation:
a. Compute a 95% CI for the true average porosity of a certain seam if the average porosity for 20 specimens from the seam was 4.85.
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of .
So it is z with a pvalue of , so
Now, find the margin of error M as such
In which is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 4.85 - 0.34 = 4.51
The upper end of the interval is the sample mean added to M. So it is 4.35 + 0.34 = 5.19
The 95% CI for the true average porosity is (4.51, 5.19).
b. Compute a 98% CI for true average porosity of another seam based on 16 specimens with a sample average of 4.56.
Following the same logic as a.
98% C.I., so
4.56 - 0.45 = 4.11
4.56 + 0.45 = 5.01
The 98% CI for true average porosity is (4.11, 5.01)
c. How large a sample size is necessary if the width of the 95% interval is to be 0.40?
A sample size of n is needed.
n is found when M = 0.4.
95% C.I., so Z = 1.96.
Rounding up
A sample size of 15 is needed.
d. What sample size is necessary to estimate the true average porosity to within 0.2 with 99% confidence?
99% C.I., so z = 2.575
n when M = 0.2.
Rounding up
A sample size of 101 is needed.