PLEASE SOLVE AND CHECK
1 answer:
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Answer:
a) The 95% confidence level two-tail confidence interval for the mean value of the key index of this batch is between 98.22 and 99.78
b) The minimum sample size to achieve this is 246.
Step-by-step explanation:
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(square root of the variance) and n is the size of the sample. So in this question,
The lower end of the interval is the sample mean subtracted by M. So it is 99 - 0.78 = 98.22
The upper end of the interval is the sample mean added to M. So it is 99 + 0.78 = 99.78.
a) The 95% confidence level two-tail confidence interval for the mean value of the key index of this batch is between 98.22 and 99.78
(b) (5 points) If we want the sampling error to be no greater than 0.5, what is the minimum sample size to achieve this based on the same confidence level with part (a)
We need a sample size of n
n is found when
Then
Rounding up
The minimum sample size to achieve this is 246.