Can someone help me answer these two questions
1 answer:
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- The margin of error of the proportion given is of 2.9%.
- Applying the margin of error, the confidence interval is (29.1%, 34.9%).
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The margin of error of a confidence interval of a proportion in a sample of size n, with a confidence level of
, is:
In which z is the z-score that has a p-value of .
The confidence interval is:
In this problem:
- Poll of 1012 people, thus
.
- 32% keep a dog, thus
95% confidence level
Thus , z is the z-score that has a p-value of
, so
.
The margin of error is:
As a percent, 2.9%, as 0.029 x 100% = 2.9%.
Now for the confidence interval, which is <u>percentage plus/minus margin of error</u>, thus:
The confidence interval is (29.1%, 34.9%).
A similar problem is given at brainly.com/question/16807970