Respuesta :
Answer:
[tex]8.2-2.58\frac{2.2}{\sqrt{18}}=6.86[/tex]
[tex]8.2+2.58\frac{2.2}{\sqrt{18}}=9.54[/tex]
So on this case the 90% confidence interval would be given by (6.86;9.54) And the error is given by:
[tex] ME= 2.58\frac{2.2}{\sqrt{18}} =1.338[/tex]
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=8.2[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma=2.2[/tex] represent the population 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 \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/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: "=-NORM.INV(0.005,0,1)".And we see that [tex]z_{\alpha/2}=2.58[/tex]
Now we have everything in order to replace into formula (1):
[tex]8.2-2.58\frac{2.2}{\sqrt{18}}=6.86[/tex]
[tex]8.2+2.58\frac{2.2}{\sqrt{18}}=9.54[/tex]
So on this case the 90% confidence interval would be given by (6.86;9.54) And the error is given by:
[tex] ME= 2.58\frac{2.2}{\sqrt{18}} =1.338[/tex]
Answer:
90% confidence interval for the population mean time = [7.944 , 8.456]
Step-by-step explanation:
We are given that the Bureau surveys 200 people. The sample mean is 8.2 minutes. There is a known standard deviation of 2.2 minutes.
Now, the pivotal quantity for 90% confidence interval for the population mean time to complete the forms is;
P.Q. = [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} }}[/tex] ~ N(0,1)
where, Xbar = sample mean = 8.2 minutes
[tex]\sigma[/tex] = population standard deviation = 2.2 minutes
n = sample size = 200
So, 90% confidence interval for the population mean time, [tex]\mu[/tex] is ;
P(-1.6449 < N(0,1) < 1.6449) = 0.90
P(-1.6449 < [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} }}[/tex] < 1.6449) = 0.90
P(-1.6449 * [tex]{\frac{\sigma}{\sqrt{n} }}[/tex] < [tex]{Xbar-\mu}[/tex] < 1.6449 * [tex]{\frac{\sigma}{\sqrt{n} }}[/tex] ) = 0.90
P(Xbar - 1.6449 * [tex]{\frac{\sigma}{\sqrt{n} }}[/tex] < [tex]\mu[/tex] < Xbar + 1.6449 * [tex]{\frac{\sigma}{\sqrt{n} }}[/tex] ) = 0.90
90% confidence interval for [tex]\mu[/tex] = [Xbar - 1.6449 * [tex]{\frac{\sigma}{\sqrt{n} }}[/tex] , Xbar + 1.6449 * [tex]{\frac{\sigma}{\sqrt{n} }}[/tex] ]
= [8.2 - 1.6449 * [tex]{\frac{2.2}{\sqrt{200} }}[/tex] , 8.2 + 1.6449 * [tex]{\frac{2.2}{\sqrt{200} }}[/tex] ]
= [7.944 , 8.456]
Therefore, 90% confidence interval for the population mean time to complete the forms is [7.944 , 8.456] .
