In a random survey of 300 self-employed individuals, 18 report having had their tax returns audited...

In a random survey of 300 self-employed individuals, 18 report having had their tax returns audited by the IRS in the past year. Construct a 98% confidence interval to estimate the proportion of self-employed individuals nationwide who've been audited by the IRS. Please round your z or t value and your final answers to 2 decimal places.

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Expert Solution

Solution :

Given that,

Point estimate = sample proportion = = x / n = 18 / 300 = 0.06

1 - = 1 - 0.06 = 0.94

Z/2 = 2.33

Margin of error = E = Z / 2 * $\sqrt$ (( * (1 - )) / n)

= 2.33 * ( $\sqrt$ ((0.06 * 0.94) / 300)

Margin of error = E = 0.03

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.06 - 0.03 < p < 0.06 + 0.03

0.03 < p < 0.09

The 98% confidence interval for the population proportion p is : 0.03 , 0.09

orchestra answered 1 year ago

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