Respuesta :
Answer:
The 90% confidence interval would be given by (6.022;6.986)
Step-by-step explanation:
1) 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 =6.504[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s=5.584 represent the sample standard deviation
n=361 represent the sample size
2) Confidence interval
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 mean and the sample deviation we can use the following formulas:
[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)
[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)
The mean calculated for this case is [tex]\bar X=6.504[/tex]
The sample deviation calculated [tex]s=5.584[/tex]
Since the sample size is large enough >30 we can use the z distirbution as approximation of the t distribution.
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=1-0.9=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that [tex]Z_{\alpha/2}=\pm 1.64[/tex]
Now we have everything in order to replace into formula (1):
[tex]6.504-1.64\frac{5.584}{\sqrt{361}}=6.022[/tex]
[tex]6.504+1.64\frac{5.584}{\sqrt{361}}=6.986[/tex]
So on this case the 90% confidence interval would be given by (6.022;6.986)
