Question

AARP reported on a study conducted to learn how long it takes individuals to prepare their federal income tax return (AARP Bulletin, April 2008). The data contained in the WEBfile named TaxReturn are consistent with the study results. These data provide the time in hours required for 40 individuals to complete their federal income tax returns. Using past years' data, the population standard deviation can be assumed known with σ = 11 hours.What is the 95% confidence interval estimate of the mean time it takes an individual to complete a federal income tax return? Round your answer to one decimal place.____ to ___ hoursReturn Preparation Time(Hours)35.330.537.426.513.049.928.844.061.60.540.534.947.936.624.139.847.818.536.639.214.537.340.549.345.528.319.55.652.641.445.339.033.729.414.540.133.736.95.633.7

Answer 1

Answer:

Confidence interval: (30.1,36.9)

Step-by-step explanation:

We are given the following data set:

35.3,30.5,37.4,26.5,13.0,49.9,28.8,44.0,61.6,0.5,40.5,34.9,47.9,36.6,24.1,39.8,47.8,18.5,36.6,39.2,14.5,37.3,40.5,49.3,45.5,28.3,19.5,5.6,52.6,41.4,45.3,39.0,33.7,29.4,14.5,40.1,33.7,36.9,5.6,33.7

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$  

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.  

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{1339.9}{40} = 33.5$

$S.D = 11$(given)

n = 40

Confidence interval:

$\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

$33.5 \pm 1.96(\frac{11}{\sqrt{40}} ) = 6.34 \pm 3.4 = (30.1,36.9)$