In a simple random sample of 35 college graduates who took a statistics course i
ID: 3133835 • Letter: I
Question
In a simple random sample of 35 college graduates who took a statistics course in college, the mean starting salary was found to be $53,000. Assume the standard deviation among all college graduates who took a statistics course is $13,000.
A. Using 99% confidence, estimate the mean starting salary among college graduates that took a statistics course.
B. What is the point estimate? What is the margin of error?
C. Explain what you mean by 99% confidence to someone that does not know much about statistical inference.
D. Suppose that the margin of error is too large, how large a sample would be required to reduce the margin of error to no more than $2,000?
Explanation / Answer
Confidence interval and sample size determination
In a simple random sample of 35 college graduates who took a statistics course in college, the mean starting salary was found to be $53,000. Assume the standard deviation among all college graduates who took a statistics course is $13,000.
A. Using 99% confidence, estimate the mean starting salary among college graduates that took a statistics course.
Solution:
The 99% confidence interval for the mean starting salary among college graduates that took a statistics course is given as below:
Confidence Interval Estimate for the Mean
Data
Population Standard Deviation
13000
Sample Mean
53000
Sample Size
35
Confidence Level
99%
Intermediate Calculations
Standard Error of the Mean
2197.4011
Z Value
-2.5758
Interval Half Width
5660.1300
Confidence Interval
Interval Lower Limit
47339.87
Interval Upper Limit
58660.13
B. What is the point estimate? What is the margin of error?
Solution:
Point estimate for the population mean is given as the sample mean = 53000 and the margin of error is given as
Margin of error = critical value * standard error
Margin of error = 2.5758*2197.4011 = 5660.13
C. Explain what you mean by 99% confidence to someone that does not know much about statistical inference.
Solution:
We are 99% sure that the average starting salary among all college graduates who took a statistics course would lies between two values such as 47339.87 and 58660.13.
D. Suppose that the margin of error is too large, how large a sample would be required to reduce the margin of error to no more than $2,000?
Solution:
Here, we have to find the sample size for the given margin of error not more than $2000. The required sample size is given as below:
Sample Size Determination
Data
Population Standard Deviation
13000
Sampling Error
2000
Confidence Level
99%
Intermediate Calculations
Z Value
-2.5758
Calculated Sample Size
280.3244
Result
Sample Size Needed
281
Confidence Interval Estimate for the Mean
Data
Population Standard Deviation
13000
Sample Mean
53000
Sample Size
35
Confidence Level
99%
Intermediate Calculations
Standard Error of the Mean
2197.4011
Z Value
-2.5758
Interval Half Width
5660.1300
Confidence Interval
Interval Lower Limit
47339.87
Interval Upper Limit
58660.13