Respuesta :
Answer:
[tex]19.1-2.306\frac{1.5}{\sqrt{9}}=17.947[/tex]
[tex]19.1+2.306\frac{1.5}{\sqrt{9}}=20.253[/tex]
So on this case the 95% confidence interval would be given by (17.947;20.253)
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=19.1[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s=1.5 represent the sample standard deviation
n=9 represent the sample size
Solution to the problem
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)
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=9-1=8[/tex]
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: "=-T.INV(0.025,8)".And we see that [tex]t_{\alpha/2}=2.306[/tex]
Now we have everything in order to replace into formula (1):
[tex]19.1-2.306\frac{1.5}{\sqrt{9}}=17.947[/tex]
[tex]19.1+2.306\frac{1.5}{\sqrt{9}}=20.253[/tex]
So on this case the 95% confidence interval would be given by (17.947;20.253)
Answer:
The 95% confidence interval for the population mean is between 15.641 years and 22.559 years.
Step-by-step explanation:
We are in posession of the sample's standard deviation, so we use the Students t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 9 - 1 = 8
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 8 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.95}{2} = 0.975([tex]t_{975}[/tex]). So we have T = 2.3060
The margin of error is:
M = T*s = 2.3060*1.5 = 3.459
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 19.1 - 3.459 = 15.641 years
The upper end of the interval is the sample mean added to M. So it is 19.1 + 3.459 = 22.559 years
The 95% confidence interval for the population mean is between 15.641 years and 22.559 years.
