Respuesta :
Answer:
The new sample size required in order to have the same confidence 95% and reduce the margin of erro to $60 is:
n=28
Step-by-step explanation:
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".
Assuming the X follows a normal distribution
[tex]X \sim N(\mu, \sigma=160)[/tex]
And the distribution for [tex]\bar X[/tex] is:
[tex]\bar X \sim N(\mu, \frac{160}{\sqrt{n}})[/tex]
We know that the margin of error for a confidence interval is given by:
[tex]Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
The next step would be find the value of [tex]\z_{\alpha/2}[/tex], [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex]
Using the normal standard table, excel or a calculator we see that:
[tex]z_{\alpha/2}=\pm 1.96[/tex]
If we solve for n from formula (1) we got:
[tex]\sqrt{n}=\frac{z_{\alpha/2} \sigma}{Me}[/tex]
[tex]n=(\frac{z_{\alpha/2} \sigma}{Me})^2[/tex]
And we have everything to replace into the formula:
[tex]n=(\frac{1.96(160)}{78.4})^2 =16[/tex]
And this value agrees with the sample size given.
For the case of the problem we ar einterested on Me= $60, and we need to find the new sample size required to mantain the confidence level at 95%. We know that n is given by this formula:
[tex]n=(\frac{z_{\alpha/2} \sigma}{Me})^2[/tex]
And now we can replace the new value of Me and see what we got, like this:
[tex]n=(\frac{1.96*160}{60})^2 =27.32[/tex]
And if we round up the answer we see that the value of n to ensure the margin of error required [tex]Me=\pm 60[/tex] $ is n=28.
