A random telephone survey of 1021 adults (aged 18 and older) was conducted by Opinion Research Corporation on behalf of Complete
Tax, an online tax preparation and e-filing service. The survey results showed that 684 of those surveyed planned to file their taxes electronically (Complete Tax Tax Prep Survey 2010). a. Develop a descriptive statistic that can be used to estimate the percentage of all taxpayers who file electronically. b. The survey reported that the most frequently used method for preparing the tax return is to hire an accountant or professional tax preparer. If 60% of the people surveyed had their tax return prepared this way, how many people used an accountant or professional tax preparer? c. Other methods that the person filing the return often used include manual preparation, use of an online tax service, and use of a software tax program. Would the data for the method for preparing the tax return be considered categorical or quantitative?
1 answer:
Answer:
a. See attachment below
b. 613 people
c. Categorical
Step-by-step explanation:
a. View attachment below
b.
Given
Total Number of People Surveyed = 1021
X tax return by professional / Number of People Surveyed = 60%
X tax /1021 = 0.6
X tax = 0.6 * 1021
X tax = 612.6
X tax = 613------ Approximated
So, the number of people used an accountant or professional tax preparer are 613.
c.
The data for preparing a tax return will be considered as categorical as the methods for a person to file his/her tax returns are categorized into four which are:
1.Electronically
2.Manual preparation
3.Online tax service
4.Use of software tax program
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I got D.
There's a few ways to solve it; I prefer using tables, but there are functions on a TI-84 that'll do it for you too. The logic here is, you have a standard normal distribution which means right away, the mean is 0 and the standard deviation is 1. This means you can use a Z table that helps you calculate the area beneath a normal curve for a range of values. Here, your two Z scores are -1.21 and .84. You might notice that this table doesn't account for negative values, but the cool thing about a normal distribution is that we can assume symmetry, so you can just look for 1.21 and call it good. The actual calculation here is:
1 - Z-score of 1.21 - Z-score .84 ... use the table or calculator
1 - .1131 - .2005 = .6864
Because this table calculates areas to the RIGHT of the mean, you have to play around with it a little to get the bit in the middle that your graph asks for. You subtract from 1 to make sure you're getting the area in the middle and not the area of the tails in this problem.
