In: Statistics and Probability
The Digest of Educational Statistics reported that the mean salary for teachers in 2012 was $56,410. Assume that this information is based on a sample of 101 teachers and the sample standard deviation was $15,000.
Does this situation meet the conditions for the t-interval of the population mean?
What is the df (degrees of freedom) for this situation?
What is the critical t-value you would use for a 90% confidence interval? (table)
What is the calculated margin of error?
What is the 90% confidence interval?
Solution :
Given that,
Yes.using the t-interval
Point estimate = sample mean = = 56410
sample standard deviation = s = 15000
sample size = n = 101
Degrees of freedom = df = n - 1 = 101 -1 =100
At 90% confidence level
= 1-0.90% =1-0.9 =0.10
/2
=0.10/ 2= 0.05
t/2,df
= t0.05,100 = 1.66
Critical value = t /2,df = 1.66
Margin of error = E = t/2,df * (s /n)
= 1.66 * ( 15000/ 101)
Margin of error = E =2477.99
The 90% confidence interval estimate of the population mean is,
- E < < + E
56410 -2477.99 < < 56410 +2477.99
53932.01 < < 58887.99.
(53932.01,58887.99.)