Question

(6 points) Elderly drivers. A polling agency interviews 748 American adults and finds that 464 think licensed drivers should be required to retake their road test once they reach 65 years of age. Round all answers to 4 decimal places. 1. Calculate the point estimate for the proportion of American adults that think licensed drivers should be required to retake their road test once they reach 65 years of age. 0.6203 2. Calculate the standard error for the point estimate you calculated in part 1. 0.0177 3. Calculate the margin of error for a 90% confidence interval for the proportion of American adults that think licensed drivers should be required to retake their road test once they reach 65 years of age. 0.0291 4. What are the lower and upper limits for the 90% confidence interval. ( 0.5912 , 0.6494 ) 5. Use the information from the polling agency to determine the sample size needed to construct a 90% confidence interval with a margin of error of no more than 3.2%. For consistency, use the reported sample proportion for the planning value of p* (rounded to 4 decimal places) and round your Z* value to 3 decimal places. Your answer should be an integer. 622.407

Answer 1

Answer:

1) p-hat = 0.6203

2) std error = 0.0177

3) margin of error = 0.0291

4) (0.5912, 0.6494)

5) minimum sample size = 1683

Step-by-step explanation:

1.  The point estimate for the proportion of American adults that think licensed drivers should be required to retake their road test once they reach 65 years of age can be calculating dividing the number of people who answer Yes by the total of people interviewed:

$\hat p=X/N=464/748=0.6203$

2. The standard error of a proportion can be expressed as:

$se_p=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.6203*0.3797}{748}}=\sqrt{0.000314877}=0.0177$

3. The z-value for a 90% CI is z=1.64485.

The margin of error can be expressed as:

$me=z\cdot se_p = 1.64485*0.0177=0.0291$

4. The lower and upper limits can be calculated as:

$LL=\hat p - z*se_p=0.6203-0.0291=0.5912\\\\\\UL=\hat p + z*se_p=0.6203+0.0291=0.6494$

5. To have a margin of error of no more than 3.2% of the sample proportion, we have to first calculate what margin of error gives 3.2% of the proportion.

$me=0.032*\hat p=0.032*0.6203=0.0198$

This margin of error can be related to the sample size as:

$me=z*se_p=z*\sqrt{\frac{p(1-p)}{n}} =1.645\sqrt{\frac{0.6203*0.3797}{n}}=0.0198\\\\1.645\sqrt{\frac{0.2355}{n} }=0.0198\\\\ \frac{0.2355}{n} =(0.0198/1.645)^2=0.0120^2=0.00014\\\\n=0.2355/0.00014\\\\n=1682.14\approx1683$