Question

(Using Excel)  The length of time for a teller to serve a customer in a bank is important. The teller service time has been recorded by camera and follows a normal distribution but has an unknown standard deviation. A simple random sample of 20 service times is selected and the times are measured. The sample mean is found to be 120 seconds. The sample standard deviation is found to be 22 seconds.  Since we do not know the population standard deviation, we will use the t distribution.

1. What are the degrees of freedom?

2. What is the standard error of xbar (s / sqrt(n))?  

3. For a 95% confidence, what is the alpha value?

4. For a 95% confidence and the degrees of freedom, what is the t-value?

5. What is the margin of error?

6. Using the 95% confidence, what is the lower confidence interval number?

7. What is the higher confidence interval number at 90% confidence?

Answer 1

#1. What are the degrees of freedom?

Ans 19

#2. What is the standard error of xbar (s / sqrt(n))?

Ans:  standard error=se=22/20^0.5==4.919

3. For a 95% confidence, what is the alpha value?

Ans  =1-c%=1-0.95=0.05

=0.05

4. For a 95% confidence and the degrees of freedom, what is the t-value?

Ans: t-value=t0.05/2,19=2.093024

5. What is the margin of error?

Ans:  

Margin of error =t*s/sqrt(n)

10.296317

6. Using the 95% confidence, what is the lower confidence interval number?

Ans:

LCL=xbar-ME

109.70368

Mean(x)=xbar=sum(x)/n	120
standard deviation(s)=sum(x-xbar)^2/n-1	22
n	20
for 95 % confidence level with degree of freedom (n-1)=19
	0.05
degrres of freedom	19
t	2.093024
se=s/sqrt(n)	4.9193496
Margin of error =t*s/sqrt(n)	10.296317
lCL=xbar+ME	130.29632
LCL=xbar-ME	109.70368