A survey shows that college students spend on average about $20 per month on Yum Good Ramen. a. Assume that the survey was conducted on 200 students and assume a standard deviation of $6. Find the 95% confidence interval of the estimate. b. Assume that the survey was conducted on 200 students and assume a standard deviation of $3. Find the 95% confidence interval of the estimate. c. Assume that the survey was conducted on 400 students and assume a standard deviation of $6. Find the 95% confidence interval of the estimate.

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[/tex] represent the sample mean

[tex]\mu[/tex] population mean (variable of interest)

[tex]\sigma[/tex] represent the population standard deviation

n represent the sample size

Part a

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 0.95 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]20-1.96\frac{6}{\sqrt{200}}=19.168[/tex]

[tex]20+1.96\frac{6}{\sqrt{200}}=20.832[/tex]

So on this case the 95% confidence interval would be given by (19.168;20.832)

Part b

For this case we just need to change the value for the deviation

[tex]20-1.96\frac{3}{\sqrt{200}}=19.584[/tex]

[tex]20+1.96\frac{3}{\sqrt{200}}=20.416[/tex]

So on this case the 95% confidence interval would be given by (19.182;20.416)

Part c

For this case we change again the deviation and the sample size.

[tex]20-1.96\frac{6}{\sqrt{400}}=19.412[/tex]

[tex]20+1.96\frac{6}{\sqrt{400}}=20.588[/tex]

So on this case the 95% confidence interval would be given by (19.412;20.588)