Question

A survey was given to a random sample of 1350 residents of a town to determine whether they support a new plan to raise taxes in order to increase education spending. Of those surveyed, 26% of the people said they were in favor of the plan. Determine a 95% confidence interval for the percentage of people who favor the tax plan, rounding values to the nearest tenth.

Answer 1

Answer:

The 95% confidence interval for the percentage of people who favor the tax plan is (23.7%, 28.3%).

Step-by-step explanation:

Confidence interval for the proportion:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

$n = 1350, \pi = 0.26$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.26 - 1.96\sqrt{\frac{0.26*0.74}{1350}} = 0.237$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.26 + 1.96\sqrt{\frac{0.26*0.74}{1350}} = 0.283$

For the percentage:

Multiply the proportions by 100%.

0.237*100 = 23.7%.

0.283*100 = 28.3%.

The 95% confidence interval for the percentage of people who favor the tax plan is (23.7%, 28.3%).

Answer 2

Answer:

$0.26 - 1.96\sqrt{\frac{0.26(1-0.26)}{1350}}=0.237$

$0.26 + 1.96\sqrt{\frac{0.26(1-0.26)}{1350}}=0.283$

Rounded the 95% confidence interval would be given by (0.2;0.3)

Step-by-step explanation:

We  know that the estimated proportion of people said they were in favor of the plan is $\hat p=0.26$

We want to construct a confidence interval with 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$. And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the true proportion of interest is given by the following formula:  

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

Replacing the info given we got:

$0.26 - 1.96\sqrt{\frac{0.26(1-0.26)}{1350}}=0.237$

$0.26 + 1.96\sqrt{\frac{0.26(1-0.26)}{1350}}=0.283$

Rounded the 95% confidence interval would be given by (0.2;0.3)