Question

A random sample of 37 taxpayers claimed an average of $9,557 in medical expenses for the year. Assume the population standard deviation for these deductions was $2,409. Construct confidence intervals to estimate average deduction for the population with the levels of significance shown below the a. 2% b. 5% c. 10% a The confidence interval with a 2% level of significance has a lower limit of S and an upper limit of Round to the nearest dollar as needed)

Answer 1

UCL = Average + z*standard deviation/(sqrt(sample size)

LCL = Average - z*standard deviation/(sqrt(sample size)

Average = 9557

standard deviation = 2409

sample size = 37

a) For 2% significance level, Service level is 98%

z value for 98% service level using z table is 2.05

UCL = 9557 + 2.05*2409/sqrt(37) = 10370

LCL = 9557 - 2.05*2409/sqrt(37) = 8744

b)

For 5% significance level, Service level is 95%

z value for 95% service level using z table is 1.64

UCL = 9557 + 1.64*2409/sqrt(37) = 10208

LCL = 9557 - 1.64*2409/sqrt(37) = 8906

c)

For 10% significance level, Service level is 90%

z value for 90% service level using z table is 1.28

UCL = 9557 + 1.28*2409/sqrt(37) = 10065

LCL = 9557 - 1.28*2409/sqrt(37) = 9049