9.3 Confidence interval for a population proportion  (Page 2/7)

Using a function of the ti-83, 83+ or 84 calculators:

• You can use technology to directly calculate the confidence interval.
• The first solution is step-by-step (Solution A).
• The second solution uses a function of the TI-83, 83+ or 84 calculators (Solution B).

Solution a

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

$q\text{'}=1-p\text{'}=1-0.600=0.400$

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 TI-83, 83+ or 84+ calculator command invNorm(0.95,0,1) 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. This can also be found using appropriate commands on other calculators, using a computer, or using a Standard Normal probability table.

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

$p\text{'}-\text{EBP}=0.60-0.036=0.564$

$p\text{'}+\text{EBP}=0.60+0.036=0.636$

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

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