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

Answer:

$ME=1.998 \frac{4}{\sqrt{64}}=0.999 \approx 1$

Step-by-step 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:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

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

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

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

$ME=t_{\alpha/2}\frac{s}{\sqrt{n}}$

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 :

$ME=1.998 \frac{4}{\sqrt{64}}=0.999 \approx 1$