Using the z-distribution, it is found that a sample size of 1066 is required.
In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.
In which z is the z-score that has a p-value of .
The margin of error is:
From the previous study, the estimate is of 0.23, hence .
98% confidence level, hence, z is the value of Z that has a p-value of , so .
The sample size is <u>n for which M = 0.03</u>, hence:
Rounding up, a sample size of 1066 is required.
A similar problem is given at brainly.com/question/12517818
Yes....one is an enlargement of the other!
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Answer:
$4,352 + $126.50= $4,478.50
Step-by-step explanation:
8.50 x 512 = 4,352
11.50 x 7 = 126.50
All together it equals
4,478.50
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Answer:
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muiplaiclthhuj mujammajan id