Question

It was reported that 59 % of individual tax returns were filed electronically in 2012. A random sample of 225 tax returns from 2013 was selected. From this sample, 142 were filed electronically.

a) 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 ______ and an upper limit of ________ . (Round to three decimal places as needed.)

***Please show all work

Answer #1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 142 / 225 = 0.631

Z_{/2}
= 1.645

Margin of error = E = Z_{
/ 2} *
((
* (1 -
)) / n)

= 1.645 * (((0.631 * 0.369) / 225)

= 0.053

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.631 - 0.053 < p < 0.631 + 0.053

0.578 < p < 0.684

Lower limit = 0.578

Upper limit = 0.684

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