You take a simple random sample of 700 adults in California (population about 40 million people) and another simple random sample of 700 adults in Wyoming (population about 578,000 people). You calculate 95% confidence intervals for the proportion in both states who ate turkey on Thanksgiving Day.
The margin of error in Wyoming ______.

The amount of random sampling error which is expressed statistically as a result of a survey is called Margin of error. Higher the margin of error, confidence of someone on the survey is lower.

The Margin of error in Wyoming is 1.6136

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Cricetus Cricetus · Answer 2 · 2022-03-04T07:35:50+01:00

The margin of error in Wyoming will be "1.6136 %".

Probability:

Probability seems to be a field of mathematics concerned mostly with determining estimated chance of occurrences of a particular phenomenon, which would be stated as nothing more than a range of 1 (One) as well as 0 (Zero).

According to the question,

Sample size, n = 700

Probability, P = 95%

Population size, N = 578000

As we know the formula,

Margin of error will be:

= 1.96 × [tex]\sqrt{\frac{N-n}{n-1} }\times \sqrt{\frac{p(1-p)}{n} }[/tex]

By substituting the values,

= 1.96 × [tex]\sqrt{\frac{578000-700}{700-1} }\times \sqrt{\frac{0.95\times (1-0.95)}{700} }[/tex] × 100

= 1.96 × [tex]\sqrt{\frac{577300}{699} }\times \frac{0.95\times 0.05}{700}[/tex] × 100

= 1.96 × [tex]\sqrt{825.89413447783}[/tex] × [tex]\sqrt{\frac{0.0475}{700} }[/tex] × 100

= 1.96 × [tex]\sqrt{825.89413447783}[/tex] × [tex]\sqrt{0.0001}[/tex] × 100

= 1.96 × 28.7384 × 0.01 × 100

= 1.6136 %

Thus the above response is correct.

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