At confidence, what is the margin of error and the interval estimate of the number of -year-old drivers in 1983? Round your intermediate answers to four decimal places. Margin of error .0190 (four decimal places) Interval estimate (to four decimal places)

The data in the question is incomplete, however I believe you are interested in knowing how to calculate margin of error ME and confidence interval CI given figures from a sampled experiment or research. The explanation box will guide you through.

Explanation:

Confidence interval CI is the range within which an outcome is expected to occur. Usually, it is tied to a percentage which suggest how sure you are that a particular outcome will occur.

Margin of error ME depicts the actual boundaries within which your prediction is permissible. Hence ME is denoted as ± thereby lying on both side of the bell curve.

The relationship between CI and ME is:

± ME and ME=  

M=sample mean

D=standard deviation

n=sample size

Z=confidence coefficient

Example:

At 98% confidence level, what is the margin of error and confidence interval estimate that from a sample size of 500 girls, the average weight is 40kg with a standard deviation of 20.

Solution:

From the question, we have the following data:

M=40kg

D=20

n=500

at 98% confidence level,

Z=2.33

because,

Z is the equivalent of p in a standard confidence coefficient table.

where p=1-

but αconfidence level=1-98%=0.02

∵p=1-=1-0.01=0.99

reading up .99 from the Z table we have

Z=2.33

now

ME==

∵Margin of error=2.08

CI=M±ME=40±2.08

Lower bound=40-2.08=37.92

Upper bound=40+2.08=42.08

In summary, it means that the researcher is 98% confident that the average weight of the girls falls withing 37.92kg to 42.08kg

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