A random sample of 49 lunch customers was taken at a restaurant. The average amo

Question

A random sample of 49 lunch customers was taken at a restaurant. The average amount of time the customers in the sample stayed in he restaurant was 33 minutes. From past experience, it is known hat the population standard deviation equals 10 minutes.

1)Compute the standard error of the mean?

2)With .99 probability compute the margin of error?

3)Construct a 99% confidence interval for the true average amount of time customers spent in the restaurant?

4)With 0.99 probability, how large of a sample would have to be taken to provide a margin of error of 2.05 minutes or less?

Explanation / Answer

a.
Standard Error= sd/ Sqrt(n)
Where,
sd = Standard Deviation
n = Sample Size

Standard deviation( sd )=10
Sample Size(n)=49
Standard Error = ( 10/ Sqrt ( 49) )
= 1.429

b.
Margin of Error = Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
Mean(x)=33
Standard deviation( sd )=10
Sample Size(n)=49
Margin of Error = Z a/2 * 10/ Sqrt ( 49)
= 2.58 * (1.429)
= 3.686

c.
Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=33
Standard deviation( sd )=10
Sample Size(n)=49
Confidence Interval = [ 33 ± Z a/2 ( 10/ Sqrt ( 49) ) ]
= [ 33 - 2.58 * (1.429) , 33 + 2.58 * (1.429) ]
= [ 29.314,36.686 ]

d.
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.01% LOS is = 2.58 ( From Standard Normal Table )
Standard Deviation ( S.D) = 10
ME =2.05
n = ( 2.58*10/2.05) ^2
= (25.8/2.05 ) ^2
= 158.391 ~ 159      

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.