Question
In: Statistics and Probability
Student's t Distribution CI The manager of The Cheesecake Factory in Boston would like to know...
Student's t Distribution CI
The manager of The Cheesecake Factory in Boston would like to know the mean number of customers served on weekdays. She believes that the number of customers served follows a Normal distribution. She reports that on six randomly selected weekdays an average of 158.3 customers were served with a sample standard deviation of 52.2.
a. Calculate the standard error of the sample mean.
b Find the t-critical value for a 99% confidence interval with a sample size of 6.
c. Calculate a 99% confidence interval of the mean.
d. Interpret the confidence interval in the context of the problem.
Solutions
Expert Solution
Solution :
Given that,
Point estimate = sample mean = 
= 158.3
sample standard deviation = s = 52.2
sample size = n = 6
Degrees of freedom = df = n - 1 = 6 - 1 = 5
a) SE = (s /
b) At 99% confidence level
= 1 - 99%
=1 - 0.99 =0.01
/2
= 0.005
t/2,df
= t0.005,5 = 4.032
c) Margin of error = E = t
/2,df
* SE
= 4.032 * 21.31
Margin of error = E = 85.92
The 99% confidence interval estimate of the population mean is,
±
E
= 158.3 ± 85.92
= ( 72.38, 244.22 )
d) We are 99% confident that the true mean of customers served on weekdaysbetween 72.38 and 244.22.
orchestra answered 2 years agoRelated Solutions
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