Question

a random sample of 233 tax returns from 2018 was selected. from this sample, 194 were filed electronically. complete parts a and b

a. construct a 99% confidence interval to estimate the actual proportion of taxpayers who filed electronically in 2018.
the confidence interval is from
(round to three decimal places as needed.)
b. what is the margin of error for this estimate?
the margin of error is
(round to three decimal places as needed.)

Explanation:

Step1: Calculate the sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 194$ (number of successes) and $n=233$ (sample size).
$\hat{p}=\frac{194}{233}\approx0.833$

Step2: Find the critical value

For a 99% confidence interval, the significance level $\alpha = 1 - 0.99=0.01$. So $\alpha/2=0.005$. The critical value $z_{\alpha/2}$ from the standard normal distribution table is $z_{0.005} = 2.576$

Step3: Calculate the margin of error (for part b)

The formula for the margin of error $E$ for a proportion is $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$
Substitute $\hat{p}=0.833$, $n = 233$, and $z_{\alpha/2}=2.576$
$E=2.576\sqrt{\frac{0.833\times(1 - 0.833)}{233}}$
First calculate inside the square - root: $0.833\times(1 - 0.833)=0.833\times0.167 = 0.139111$
$\frac{0.139111}{233}\approx0.000597$
$\sqrt{0.000597}\approx0.0244$
$E=2.576\times0.0244\approx0.063$

Step4: Calculate the confidence interval (for part a)

The confidence interval for a proportion is $\hat{p}-E$\hat{p}-E=0.833-0.063 = 0.770$
$\hat{p}+E=0.833 + 0.063=0.896$

Answer:

a. The confidence interval is from $0.770$ to $0.896$
b. The margin of error is $0.063$