Students in a local high school decide to estimate the propo

Students in a local high school decide to estimate the proportion of the student body (size 1.500) who arc left handed. Out of a random sample of 50 students, 12% indicate they are left handed. The students obtain a 95% confidence interval for the proportion of all students at the high school are left-handed of 3% to 21%. What can we conclude about their confidence interval A The usual interpretation of the confidence interval applies: they can be 95% confident in their confidence interval. The usual interpretation of the confidence interval does not apply: they cannot be 95% confident in their confidence interval since the success/failure condition is violated. The usual interpretation of the confidence interval does not apply: they cannot be 95% confidence in their confidence interval since the randomization condition is violated. The usual interpretation of the confidence interval does not apply: they cannot be 95% confidence in their confidence interval since the 10% condition is violated. In a Gallup poll, 26.6% of 5000 people reported a Body Mass Index (BMI) greater than 30, which is classified as obese. The 5000 people polled are a random sample of adults, aged 18 and older, from the United States population. Choose the correct interpretation for the 98% confidence interval for the proportion of the population who are obese The proportion of adults in the United States, aged 18 or older, who are obese is between 0.2515 and 0.2805. We are 98% confident the proportion of adults polled who are obese is between 0.2515 and 0.2805. In 98% of all samples, the proportion of adults polled who are obese will be between 0.2515 and 0.2805. D. We are 98% confident that the proportion of adults in the United States, aged 18 or older, who arc obese is between 0.2515 and 0.2805. What sample size is necessary to obtain a 99% confidence interval for a population proportion with a margin of error of no more than 4% 601 846 10374148

Solution

1) D 1.28

2) A

3) D

Confidence interval:

(0.266 - 2.33*sqrt(0.266*(1-0.226)/5000) , 0.266 + 2.33*sqrt(0.266*(1-0.226)/5000)

= (0.2515,0.2805) Answer

4) C . 1037

M = z*sqrt(p(1-p)/n)

0.04 = 2.58 * sqrt(p*(1-p)/n)

=> 0.04/2.58 = sqrt(1/4n)

=> 0.0155 = sqrt(1/4n)

=> 1/4n = 0.00024025

=> n = 1037 Answer