Question

M118 Worksheet Creating a Confidence interval, 2 Standard Deviation Approach (A) In a survey of 1002 people, 701 said that they had voted in a recent presidential election. Create a 9596 Cl for the percentage of all people who say they voted in the election () Voting records show that 61% of registered voters actually voted in the election, Is this data consistent with your CL. -250 +25D (2) A SRS of 1228 medical malpractice lawsults showed that 856 of these sults were dismissed. Create a 95% CI for the percentage of all medical malpractice lawsuits that are dropped. Does it appear that the majority of such suits are dismissed. P -25D 2D

Answer 1

The sample proportion is computed as below,

$\hat p = \frac{X}{N} = \frac{701}{1002} = 0.6996$

The critical value for 95% confidence interval is 1.96.

The corresponding confidence interval is computed as shown below:

$CI = (\hat p - z_c \sqrt{\frac{\hat p (1-\hat p)}{n}}, \hat p + z_c \sqrt{\frac{\hat p (1-\hat p)}{n}})$

$CI = ( 0.6996 - 1.96 \times \sqrt{\frac{0.6996 (1- 0.6996)}{1002}}, 0.6996 + 1.96 \times \sqrt{\frac{0.6996 (1- 0.6996)}{1002}} )$

= $(0.6712, 0.7280)$

Therefore, the 95% confidence interval for the population proportion lies between 0.6712 < p < 0.7280.

B)

No, the data is not consistent with the Confidence interval constructed as the population proportion does not lie between the lower and upper bound between the interval.

2)

The sample proportion is computed as below,

$\hat p = \frac{X}{N} = \frac{856}{1228} = 0.6971$

The critical value for 95% confidence interval is 1.96.

The corresponding confidence interval is computed as shown below:

$CI = (\hat p - z_c \sqrt{\frac{\hat p (1-\hat p)}{n}}, \hat p &plus; z_c \sqrt{\frac{\hat p (1-\hat p)}{n}})$

$CI = ( 0.6971 - 1.96 \times \sqrt{\frac{0.6971 (1- 0.6971)}{1228}}, 0.6971 &plus; 1.96 \times \sqrt{\frac{0.6971 (1- 0.6971)}{1228}} )$

= $(0.6714, 0.7228)$

Therefore, the 95% confidence interval for the population proportion is 0.6714 < p < 0.7228.