Answer:
Step-by-step explanation:
There is <em>one-and-a-half</em> <em>hour</em> in <em>ninety</em><em> </em><em>minutes</em>,<em> </em>so multiply 6 by 15 to get 90, then whatever is done to the bottom is also done to the top, so you <em>square</em> 6 to get 36.
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Using the z-distribution, we have that:
a) A sample of 601 is needed.
b) A sample of 93 is needed.
c) A. Yes, using the additional survey information from part (b) dramatically reduces the sample size.
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:
95% confidence level, hence, z is the value of Z that has a p-value of
, so
.
For this problem, we consider that we want it to be within 4%.
Item a:
Rounding up:
A sample of 601 is needed.
Item b:
The estimate is , hence:
Rounding up:
A sample of 93 is needed.
Item c:
The closer the estimate is to , the larger the sample size needed, hence, the correct option is A.
For more on the z-distribution, you can check brainly.com/question/25404151