Question

Assume that population means are to be estimated from the samples described. Use the sample results to approximate the margin of error and​ 95% confidence interval. Sample sizeequals1 comma 037​, sample meanequals​$46 comma 236​, sample standard deviationequals​$25 comma 000

Answer 1

`ME=1.96(25000)/(√(1037))=1521.622`

Explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`\bar X=46236` represent the sample mean for the sample

`\mu` population mean (variable of interest)

s=25000 represent the sample standard deviation

n=1037 represent the sample size

2) Confidence interval

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

We can assume that the sample size is large enough to use the z distribution for the quantile and we can assume that the sample deviation is the best estimator for the population deviation
`\hat \sigma =s`

Since the Confidence is 0.95 or 95%, the value of
`\alpha=0.05` and
`\alpha/2 =0.025`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that
`z_(\alpha/2)=1.96`

Now we have everything in order to replace into formula (1):

`46236-1.96(25000)/(√(1037))=44.7143785`

`46236+1.96(25000)/(√(1037))=47757.622`

So on this case the 95% confidence interval would be given by (3229.95;3326.49)

3) Margin of error

The margin of error is given by this formula:

`ME=z_(\alpha/2)(s)/(√(n))` (2)

If we replace we got:

`ME=1.96(25000)/(√(1037))=1521.622`