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)

Answer :

Answer:

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:

[tex]CI=M[/tex]± ME and ME=[tex]Z*\frac{D}{\sqrt{n} }[/tex]  

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-[tex]\frac{\alpha }{2}[/tex]

but α[tex]=1-[/tex]confidence level=1-98%=0.02

∵p=1-[tex]\frac{0.02}{2}[/tex]=1-0.01=0.99

reading up .99 from the Z table we have

Z=2.33

now

ME=[tex]Z*\frac{D}{\sqrt{n} }[/tex]=[tex]=2.33*\frac{20}{\sqrt{500} }[/tex][tex]=2.33*0.89=2.08[/tex]

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

Other Questions