Question

A random sample of 37 taxpayers claimed an average of ​$9,527 in medical expenses for the year. Assume the population standard deviation for these deductions was ​$2,380.

Construct confidence intervals to estimate the average deduction for the population with the levels of significance shown below.

a.	5%
b.	10%
c.	20%

The confidence interval with a 5 % level of significance has a lower limit of ​_and an upper limit of _

The confidence interval with a 10% level of significance has a lower limit of ​_and an upper limit of _

The confidence interval with a 20% level of significance has a lower limit of ​_and an upper limit of _

Answer 1

a)

sample mean, xbar = 9527
sample standard deviation, σ = 2380
sample size, n = 37

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

ME = zc * σ/sqrt(n)
ME = 1.96 * 2380/sqrt(37)
ME = 766.89

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (9527 - 1.96 * 2380/sqrt(37) , 9527 + 1.96 * 2380/sqrt(37))
CI = (8760.11 , 10293.89)

lower limit of 8760.11 and an upper limit of 10293.89

2)

sample mean, xbar = 9527
sample standard deviation, σ = 2380
sample size, n = 37

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.64

ME = zc * σ/sqrt(n)
ME = 1.64 * 2380/sqrt(37)
ME = 641.68

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (9527 - 1.64 * 2380/sqrt(37) , 9527 + 1.64 * 2380/sqrt(37))
CI = (8885.32 , 10168.68)

lowerlimit of 8885.32 upper limit of 10168.68

c)

sample mean, xbar = 9527
sample standard deviation, σ = 2380
sample size, n = 37

Given CI level is 80%, hence α = 1 - 0.8 = 0.2
α/2 = 0.2/2 = 0.1, Zc = Z(α/2) = 1.28

ME = zc * σ/sqrt(n)
ME = 1.28 * 2380/sqrt(37)
ME = 500.83

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (9527 - 1.28 * 2380/sqrt(37) , 9527 + 1.28 * 2380/sqrt(37))
CI = (9026.17 , 10027.83)

lower limit of 9026.17 and upper limit of, 10027.83