Respuesta :
Answer:
[tex]11-2.385\frac{19}{\sqrt{66}}=5.422[/tex]
So on this case the one lower 99% confidence interval would be given by (5.422;11)
And the lower bound for this case would be 5.422, and that means "the minimum value for the mean at 99% confidence for the mean"
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
The confidence interval for the mean is given by the following formula:
[tex]\bar X - t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=66-1=65[/tex]
Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,65)".And we see that [tex]t_{\alpha/2}=2.385[/tex]
Now we have everything in order to replace into formula (1):
[tex]11-2.385\frac{19}{\sqrt{66}}=5.422[/tex]
So on this case the one lower 99% confidence interval would be given by (5.422;11)
And the lower bound for this case would be 5.422, and that means "the minimum value for the mean at 99% confidence for the mean"
