Respuesta :
Answer:
[tex]ME=1.96\frac{25000}{\sqrt{1037}}=1521.622[/tex]
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=46236[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s=25000 represent the sample standard deviation
n=1037 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)
We can assume that the sample size is large enough to use the z distribution for the quantile and we can assume that the sample deviation is the best estimator for the population deviation [tex]\hat \sigma =s[/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: "=-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]46236-1.96\frac{25000}{\sqrt{1037}}=44.7143785[/tex]
[tex]46236+1.96\frac{25000}{\sqrt{1037}}=47757.622[/tex]
So on this case the 95% confidence interval would be given by (3229.95;3326.49)
3) Margin of error
The margin of error is given by this formula:
[tex] ME=z_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (2)
If we replace we got:
[tex]ME=1.96\frac{25000}{\sqrt{1037}}=1521.622[/tex]
