Respuesta :
Answer:
[tex]14.42-1.96\frac{6.95}{\sqrt{36}}=12.150[/tex]
[tex]14.42+ 1.96\frac{6.95}{\sqrt{36}}=16.690[/tex]
So on this case the 95% confidence interval would be given by (12.150;16.690)
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=14.42[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma=6.95[/tex] represent the population standard deviation
n=36 represent the sample size
Solution to the problem
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], 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 [tex]z_{\alpha/2}=1.96[/tex]
Now we have everything in order to replace into formula (1):
[tex]14.42-1.96\frac{6.95}{\sqrt{36}}=12.150[/tex]
[tex]14.42+ 1.96\frac{6.95}{\sqrt{36}}=16.690[/tex]
So on this case the 95% confidence interval would be given by (12.150;16.690)
