What is 60,000+3,000+20+9 in standard form?
1 answer:
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Using the z-distribution and the formula for the margin of error, it is found that:
a) A sample size of 54 is needed.
b) A sample size of 752 is needed.
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 of:
90% confidence level, hence, z is the value of Z that has a p-value of
, so
.
Item a:
The estimate is .
The sample size is <u>n for which M = 0.03</u>, hence:
Rounding up, a sample size of 54 is needed.
Item b:
No prior estimate, hence
Rounding up, a sample of 752 should be taken.
A similar problem is given at brainly.com/question/25694087