Question

A random sample of 20 items is selected from a population. When computing a confidence interval for the population mean, what number of degrees of freedom should be used to determine the appropriate t-value?

Answer 1

Answer:

For this case since the sample size is lower than 30 , n =20<30, is not appropiate use the normal standard distribution to calculate the confidence interval for the mean, and then for this case is better use the t distribution since takes in count the correction factor to approximate the distribution of the true parameter.

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

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (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".

$\bar X$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

For this case since the sample size is lower than 30 , n =20<30, is not appropiate use the normal standard distribution to calculate the confidence interval for the mean, and then for this case is better use the t distribution since takes in count the correction factor to approximate the distribution of the true parameter.

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

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)