Question

8. A 95% confidence interval for population proportion of STA2023 students who studied more than an hour for their last exam is (45 % , 55 % ) . The margin of error of this interval is: 8% c. 5% a. d. 42% b. 10% Short Answer Section: Read each question carefully, and answer all parts using complete sentences, proper grammar, and proper statistical terminology. Answers need to be legible. 9. In a recent poll of 100 randomly selected adults in the United States, 44 believed that they had encountered a ghost in their home. What is the sample proportion? (2 points) a. b. Show the conditions for approximate normality are met (CLT). (3 points) c. What is the margin of error for a 90 % confidence interval? Show your work, and round to three places. (3 points) d. What is the 90% confidence interval for this scenario? Round to three places. (4 points) possible population proportion of U.S. adults who e. Would you support 39% as a believed that they had encountered a ghost in their home? Why or why not? (3 points)

Answer 1

8) The 95% confidence for the population proportion is (45%, 55%)

That is a lower limit of confidence interval = 45%

and upper limit of confidence interval = 55%

The formula of a margin of error using both limits is,

$Margin \ of \ error = \frac{Upper \ limit - Lower \ limit}{2} = \frac{55 - 45}{2} = \frac{10}{2} = 5$

Margin of error = 5%

Option c is correct.

9. Given:

n = number of adults selected = 100

x = number of adults who believed they had encountered a ghost in their home = 44

a) Sample proportion:

The formula to find sample proportion is,

Sample proportion() = - = 0.44 100

b) Conditions for normality.

If both are greater than or equal to 10 then the distribution of sample proportion is approximately normal according to the Central limit theorem.

np and n(1-P)

Here both are greater than or equal to 10,

np= 100 * 0.44 = 44 and n(1-7) = 100* (1 -0.44) = 56

so the distribution of phat is approximately normal.

c) The margin of error.

The formula of the margin of error is,

$Margin \ of \ error (ME) = Z_{\alpha/2}\sqrt{\frac{\hat p *(1-\hat p)}{n}}$

Z is the critical value at a given confidence level.

c = confidence level = 0.90

$\alpha = 1 - c = 1 - 0.90 = 0.10$

The area to the left of $Z_{\alpha/2}$ is

1 - (a/2) = 1 - (0.10/2) = 1 -0.05 = 0.95

The z critical value for the area 0.95 using the z table is 1.645

The margin of error is,

$ME = Z_{\alpha/2}\sqrt{\frac{\hat p *(1-\hat p)}{n}} = 1.645 \sqrt{\frac{0.44(1-0.44)}{100}} = 0.082$

Margin of error = 0.082

d) 90% confidence interval for the population proportion

The formula of a confidence interval is

$\hat p - margin \ of \ error < p < \hat p + margin \ of \ error$

0.358 < p < 0.522

The 90% confidence interval for population proportion is (0.358, 0.522)

e) Yes, 39% (0.39) falls between the confidence interval (0.358, 0.522) .

Therefore, we support 39% as a possible population proportion of U. S. adults who believed that they had encountered a ghost in their home.