Question

In: Math

A random sample of 43 taxpayers claimed an average of $9,853 in medical expenses for the...

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 ______.

Solutions

Expert Solution

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 .


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