Question
In: Statistics and Probability
A university surveyed recent graduates of the English Department for their starting salaries. Four hundred graduates...
A university surveyed recent graduates of the English Department for their starting salaries. Four hundred graduates returned the survey. The average salary was $25,000. The population standard deviation was $2,500. What is the 90% confidence interval for the mean salary of all graduates from the English Department?
A) [$24,794, $25,206]
B) [$24,988, $25,012]
C) [$22,500, $27,500]
D) [$24,755, $25,245]
Solutions
Expert Solution
Solution
Given that,
= 25000
= 2500
n = 400
At 90% confidence level the z is ,
= 1 - 90% = 1 - 0.90 = 0.10
/ 2 = 0.10 / 2 = 0.05
Z/2
= Z0.05 = 1.645
Margin of error = E = Z/2*
(
= 1.645 * (2500 /
= 206
At 90% confidence interval estimate of the population mean is,
- E < 
<
+ E
25000- 206 <
< 25000 + 206
25794<
< 25206
(25794, 25206)
Option a ) is correct.
orchestra answered 2 years agoRelated Solutions
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