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ivann1987 [24]
3 years ago
12

A survey of 132 students is selected randomly on a large university campus. They are asked if they use a laptop in class to take

notes. The result of the survey is that 66 of the 132 students responded "yes.". An approximate 98% confidence interval is (0.399, 0.601). How would the confidence interval change if the confidence level had been 90% instead of 98%
Mathematics
1 answer:
ddd [48]3 years ago
7 0

Answer:

For 90% CI = (0.428, 0.572)

For 98% CI = (0.399, 0.601)

The confidence interval (and Margin of error) reduces when 90% confidence level is used compared to when 98% confidence level is used.

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

p+/-z√(p(1-p)/n)

Given that;

Proportion p = 66/132 = 0.50

Number of samples n = 132

Confidence level = 90%

z(at 90% confidence) = 1.645

Substituting the values we have;

0.50 +/- 1.645√(0.50(1-0.50)/132)

0.50 +/- 1.645√(0.001893939393)

0.50 +/- 0.071589436011

0.50 +/- 0.072

(0.428, 0.572)

The 90% confidence level estimate of the true population proportion of students who responded "yes" is (0.428, 0.572)

For 90% CI = (0.428, 0.572)

For 98% CI = (0.399, 0.601)

The confidence interval (and Margin of error) reduces when 90% confidence level is used compared to when 98% confidence level is used.

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Answer:

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2) New York City.

Here we can think this as:

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As larger is the area of the city, more probable to being randomly choosen, so to find the exact probability we need to find the quotient between the area of New York City and the total area of North America:

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3 years ago
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