Question

Assume that population mean is to be estimated from the sample described. Use the sample results to approximate the margin of error and​ 95% confidence interval. Sample​ size, n=64​; sample​ mean, x =69.0 ​cm; sample standard​ deviation, s=4.0 cm

Answer 1

`ME=1.998 (4)/(√(64))=0.999 \approx 1`

Explanation:

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 =69` represent the sample mean for the sample

`\mu` population mean

s=4.0 represent the sample standard deviation

n=64 represent the sample size

Assuming the X follows a normal distribution

`X \sim N(\mu, \sigma)`

The sample mean
`\bar X` have the following distribution:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

The margin of error for the sample mean is given by this formula:

`ME=z_(\alpha/2)(\sigma)/(√(n))`

Since we have the sample standard deviation we can estimate the margin of error like this:

`ME=t_(\alpha/2)(s)/(√(n))`

The next step would be find the value of
`\z_(\alpha/2)`,
`\alpha=1-0.95=0.05` and
`\alpha/2=0.025`

We can find the degrees of freedom like this:

`df=n-1=64-1=63`

And we can find the critical value with the following code in excel for example: "=T.INV(0.025,63)" and we got:

`t_(\alpha/2)=\pm 1.998`

And the margin of error would be :

`ME=1.998 (4)/(√(64))=0.999 \approx 1`