Question

You have drawn a simple random sample of 36 college students, asking each student how much rent they pay per month. You obtain a sample mean of 573 and a sample standard deviation of 22. Calculate the upper confidence limit of a 90% confidence interval for the population mean of rent paid. If necessary, round your answer to four decimal places. HINT: For a confidence interval of [a,b] we refer to "a" as the "lower confidence limit" and we refer to "b" as the "upper confidence limit". g

Answer 1

Answer:

$573-1.690\frac{22}{\sqrt{36}}=566.8033$    

$573+1.690\frac{22}{\sqrt{36}}=579.1967$      

And the 90% confidence interval would be between [566.80 and 579.20]

Step-by-step explanation:

Information given

$\bar X=573$ represent the sample mean

$\mu$ population mean (variable of interest)

s=22 represent the sample standard deviation

n=36 represent the sample size  

Confidence interval

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

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

The degrees of freedom are given by:

$df=n-1=36-1=35$

The Confidence level is 0.90 or 90%, the significance would be $\alpha=0.1$ and $\alpha/2 =0.05$, and the critical value for this case would be $t_{\alpha/2}=1.690$

And replacing we got:

$573-1.690\frac{22}{\sqrt{36}}=566.8033$    

$573+1.690\frac{22}{\sqrt{36}}=579.1967$    

And the 90% confidence interval would be between [566.8033 and 579.1967]