Respuesta :
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:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (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".
[tex]\bar X[/tex] represent the sample mean for the sample
[tex]\mu[/tex] 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:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
