Question

A random sample of 40 taxpayers claimed an average of $9.920 in medical expenses for the year. Assume the population standard deviation for these deductions was 52 370. Construct confidence intervals to estimate the average deduction for the population with the levels of significance shown below. a 190 520 c. 10% a. The confidence interval with a 15. level of significance has a lower limit of s (Round to the nearest dollar as needed) and an upper limit of

Answer 1

Avergae = x̄ = $9920

Population standard deviation = σ = $2370

Random sample size = n = 40

z-score = z

Confidence interval for population mean µ is given as-

$\overline{x} \pm z\frac{\sigma }{\sqrt{n}}$

a) Here α = 1% = 0.01

Z-score Confidence Level Level of Significance (alpha) (1 - confidence level] Area in one tail (alpha/2) 50% 0.5 0.2500 0.674 80% 0.2 0.1000 1.282 90% 0.1 0.0500 1.645 95% 0.05 0.0250 1.960 98% 0.02 0.0100 2.326 99% 0.01 0.0050 2.576

z-score =z= 2.576

Confidence Interval :-

$9920 \pm 2.576\frac{2370 }{\sqrt{40}} = (8955,10885)\: [after\:rounding \:off]$

b) Here, α = 2% = 0.02

z-score = z=2.326

Confidence Interval :-

$9920 \pm 2.326\frac{2370 }{\sqrt{40}} = (9048,10792)\: [after\:rounding \:off]$

c) Here, α = 10% = 0.1

z-score = z=1.645

Confidence Interval :-

$9920 \pm 1.645\frac{2370 }{\sqrt{40}} = (9304,10536)\: [after\:rounding \:off]$

*Note, in all these three parts, rounding off is done ignoring the decimals.