It was reported that 62% of individual tax returns were filed electronically in
ID: 3158744 • Letter: I
Question
It was reported that 62% of individual tax returns were filed electronically in 2012. A random sample of 175 tax returns from 2013 was selected. From this sample. 115 were filed electronically. Complete parts a through c below. Construct a 90% confidence interval to estimate the actual proportion of taxpayers who filed electronically in 2013. A 90% confidence interval to estimate the actual proportion has a lower limit of Square and an upper limit of Square. (Round to three decimal places as needed.) What is the margin of error for this sample? The margin of error is Square. (Round to three decimal places as needed.) Is there any evidence that this proportion has changed since 2012 based on this sample? Because the confidence interval found in part a the reported proportion from 2012,this sample evidence that this proportion has changed since then.Explanation / Answer
Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=115
Sample Size(n)=175
Sample proportion = x/n =0.657
Confidence Interval = [ 0.657 ±Z a/2 ( Sqrt ( 0.657*0.343) /175)]
= [ 0.657 - 1.645* Sqrt(0.001) , 0.657 + 1.65* Sqrt(0.001) ]
= [ 0.598,0.716]
Margin of Error = Z a/2 Sqrt(p*(1-p)/n))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=115
Sample Size(n)=175
Sample proportion =0.657
Margin of Error = Z a/2 * ( Sqrt ( (0.657*0.343) /175) )
= 1.645* Sqrt(0.001)
=0.059
a. [ 0.598,0.716]
b. 0.059
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