In: Accounting
A researcher has collected a sample of 25 respondents and the mean was calculated as 500. The sample was randomly drawn from a normal population whose standard deviation is 15. a) Estimate the population mean with 99% confidence b) Repeat the part (a) changing the population standard deviation to 30. c) Repeat the part (a) changing the population standard deviation to 60. d) Describe what happens to the confidence interval estimate when the standard deviation is increased.
Sample mean = = 500
Sample size = n = 25
Confidence interval is 99%
So here 1-α=.99 (α= confidence coefficient of the interval)
α=.01
α/2 = .005
The standard error (S.E) of the sample mean= ( when population S.D ‘σ’ is known).
Let ‘t’ be the table value of the statistic for 'α' level of significance and for the desired degree of freedom.
Then the confidence limit for the population mean=±(t×S.E).
Here t=2.58
Note: Values of ‘t’ are available in the Statistical table.
a. When population standard deviation σ=15.
S.E=σ/√n =15/√25 = 15/5 =3
±(t×S.E )= 500 ± (2.58×3) = 500±7.74 =(492.26,507.74)
Therefore the population mean with 99% confidence ranges from 492.26 to 507.74
b. When population standard deviation σ=30.
S.E=σ/ √n =30/ √25 = 30/5 =6
±(t×S.E) = 500 ± (2.58×6) = 500±15.48 =(484.52,515.48)
Therefore the population mean with 99% confidence ranges from 484.52 to 515.48
c. When population standard deviation σ=60.
S.E=σ/√n =60/√25 = 60/5 =12
±(t×S.E) = 500 ± (2.58×12) = 500±30.96 =(469.04,530.96)
Therefore the population mean with 99% confidence ranges from 469.04 to 530.96
d. When the standard deviation is increased, the range of the confidence interval estimate also increases.