Question

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.

Answer 1

Answer:

a) The 95% confidence interval would be given by (19.168;20.832)

b) The 95% confidence interval would be given by (19.182;20.416)

c) The 95% confidence interval would be given by (19.412;20.588)

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".

$\bar X$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma$ represent the population standard deviation

n represent the sample size  

Part a

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, 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 $z_{\alpha/2}=1.96$

Now we have everything in order to replace into formula (1):

$20-1.96\frac{6}{\sqrt{200}}=19.168$    

$20+1.96\frac{6}{\sqrt{200}}=20.832$

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

$20-1.96\frac{3}{\sqrt{200}}=19.584$    

$20+1.96\frac{3}{\sqrt{200}}=20.416$

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.

$20-1.96\frac{6}{\sqrt{400}}=19.412$    

$20+1.96\frac{6}{\sqrt{400}}=20.588$

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