8.6 Confidence intervals: confidence interval for a population  (Page 2/7)

• Start by discussing the assumptions and conditions that support your model.
• The solution is step-by-step.

Solution a

• Randomization – no information is given in the problem about the sampling methodology.
• independence – wsince we do not know how were respondents were chosen we do not know if responses are independent.
• 10% condition – the sample size of 500 is less than 10% all adult residents in the large city.
• Success/ Failure – $\hat{p}n>10\mathrm{(1-}\hat{p})n>10$ 0.60(500) = 300>10 and 1-0.60)(500)=200>10 where $\hat{p}$ is the sample percentage and n is the sample size. Both products are larger than 10, so the success/ failure condition is met.(When assumptions and conditions are not met we note that in our work, but continue on with the problem.)

$\hat{p}=\frac{x}{n}=\frac{300}{500}=0.600$

$\hat{q}=1-\hat{p}=1-0.600=0.400$

Since $\text{CL}=0.90$ , then $\alpha =1-\text{CL}=1-0.90=0.10\frac{\alpha }{2}=0.05$ .

$z_{\frac{\alpha }{2}}=z_{.05}=1.645$

Use the the standard normal probability table to find $z_{.05}$ . Remember that the area to the right of $z_{.05}$ is 0.05 and the area to the left of $z_{.05}$ is 0.95.

$\text{ME}=z_{\frac{\alpha }{2}}\cdot \sqrt{\frac{\hat{p}\cdot \hat{q}}{n}}=1.645\cdot \sqrt{\frac{\left(0.60\right)\cdot \left(0.40\right)}{500}}=0.036$

$\hat{p}-\text{ME}=0.60-0.036=0.564$

$\hat{p}+\text{ME}=0.60+0.036=0.636$

The confidence interval for the true binomial population proportion is $(\hat{p}-\text{ME},\hat{p}+\text{ME})=$ $(0.564,0.636)$ .

Interpretation:

• We estimate with 90% confidence that the true percent of all students that are registered voters is between 56.4% and 63.6%.
• Alternate Wording: We estimate with 90% confidence that between 56.4% and 63.6% of ALL students are registered voters.

Explanation of 90% confidence level