In: Math
A random sample of 43 taxpayers claimed an average of $9,853 in medical expenses for the year. Assume the population standard deviation for these deductions was 2,418. Construct confidence intervals to estimate the average deduction for the population with the levels of significance shown below.
a.1%
b.5%
c.20%
a. The confidence interval with a 1% level of significance has a lower limit of _____ and an upper limit of ______.
b. The confidence interval with a 5% level of significance has a lower limit of _____ and an upper limit of ______.
c. The confidence interval with a 20% level of significance has a lower limit of _____ and an upper limit of ______.
Solution :
Given that,
= 9853
= 2418
n = 43
(a)
= 1% = 0.01
/ 2 = 0.01 / 2 = 0.005
Z/2 = Z0.005 = 2.576
Margin of error = E = Z/2* ( /n)
= 2.576 * (2418 / 43)
= 950
At 99% confidence interval estimate of the population mean is,
- E < < + E
9853 - 950 < < 9853 + 950
8903 < < 10803
The confidence interval with a 1% level of significance has a lower limit of 8903 and
an upper limit of 10803 .
(b)
= 5% = 0.05
/ 2 = 0.05 / 2 = 0.025
Z/2 = Z0.025 = 1.96
Margin of error = E = Z/2* ( /n)
= 1.96 * (2418 / 43)
= 723
At 95% confidence interval estimate of the population mean is,
- E < < + E
9853 - 723 < < 9853 + 723
9130 < < 10576
The confidence interval with a 5% level of significance has a lower limit of 9130 and
an upper limit of 10576 .
(c)
= 20% = 0.20
/ 2 = 0.20 / 2 = 0.10
Z/2 = Z0.10 = 1.28
Margin of error = E = Z/2* ( /n)
= 1.28 * (2418 / 43)
= 464
At 80% confidence interval estimate of the population mean is,
- E < < + E
9853 - 464 < < 9853 + 464
9389 < < 10317
The confidence interval with a 20% level of significance has a lower limit of 9389 and
an upper limit of 10317 .