Question

A study of a local high school tried to determine the mean amount of money that each student had saved. The study surveyed a random sample of 96 students in the high school and found a mean savings of 4000 dollars with a standard deviation of 500 dollars. Determine a 95% confidence interval for the mean, rounding all values to the nearest whole number.

Answer 1

I don't know sorry I will try and come back If I get it :D

Answer 2

A 95% confidence interval for the average is 4100

What is confidence interval ?

The confidence interval formula is used in statistics for describing the amount of uncertainty associated with a sample estimate of a population parameter. It is used to describe the uncertainty associated with a sampling method.

Formula of confidence interval:

If n ≥ 30,

Confidence Interval = m ± $z(\frac{SD}{\sqrt{n} } )$

where,

n = Number of terms

m = Sample average

SD = Standard Deviation

z = Value corresponding to confidence interval in z table

Table of Confidence interval  and Z

CI      z

0.70 1.04

0.75 1.15

0.80 1.28

0.85 1.44

0.90  1.645

0.92 1.75

0.95 1.96

0.96 2.05

0.98 2.33

0.99 2.58

According to the question

random sample of student (n) =  96

average savings = 4000 dollars

Standard deviation =  500 dollars

95% confidence interval for the average needed

i.e , z = 1.96

Now, By using Formula of confidence interval:

Confidence Interval = m ± $z(\frac{SD}{\sqrt{n} } )$

substituting the values

Confidence Interval = 4000 ± $1.96(\frac{500}{\sqrt{96} } )$

                                  = 4000 ± 100.021

                                  = 4100.21

values to the nearest whole number

Confidence Interval = 4100

Hence,  A 95% confidence interval for the average is 4100 .

To know more about confidence interval  here::

https://brainly.com/question/13067956

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