CAN SOMEONE PLEASE HELP ME???
1 answer:
You might be interested in
Answer:
a) The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams
b) 49 measurements are needed.
Step-by-step explanation:
Question a:
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 .
That is z with a pvalue of , so Z = 2.327.
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 10.0044 - 0.00031 = 10.00409 grams
The upper end of the interval is the sample mean added to M. So it is 10 + 0.00031 = 10.00471 grams
The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams.
(b) How many measurements must be averaged to get a margin of error of +/- 0.0001 with 98% confidence?
We have to find n for which M = 0.0001. So
Rounding up
49 measurements are needed.