In: Statistics and Probability
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.
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 * (( * (1 - )) / n)
= 2.33 * (((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