A survey of 280 homeless persons showed that 63 were veterans. Construct a 90% confidence interval for the proportion of homeless persons who are veterans. (a) (0.167, 0.283) (b) (0.176, 0.274) (c) (0.184, 0.266) (d) (0.161 0.380)
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- Suppose you take 4 air quality samples in parts per million: 19.6, 1.88, 12.7, 11.9. Give a 90% confidence interval of the mean air quality? (Also, t(0.95,3) = 2.3534 and t(0.90,4)= 1.5332) (a) (3.45, 16.83) (b) (0, 25) (c) (2.93, 20.10) (d) (5.22, 21.54) (e) None of the aboveFrom a random sample of 100 women, 15 have dangerous reading blood sugar levels higher than 300 mg/dL. The 95% confidence interval for the proportion of all women with dangerous reading levels is O a. 0.15+ (1.96) √0.1275 O b. O d. 0.15 0.15 (1.6449) √0.0013 (1.6449) √0.1275 0.15 (1.96) √0.0013A random sample of n1 278 people who live in a city were selected and 98 identified as a "dog person." A random sample of n2 = 109 people who live in a rural area were selected and 65 identified as a "dog person." Find the 90% confidence interval for the difference in the proportion of people that live in a city who identify as a "dog person" and the proportion of people that live in a rural area who identify as a "dog person." Round answers to to 4 decimal places. < pi – p2 < Submit Question
- Quesiton 20Which two of the following could be confidence intervals with the same sample? (0.45, 0.75) (0.4, 0.5) (0.35, 0.45) (0.3, 0.6)You recently sent out a survey to determine if the percentage of adults who use social media has changed from 56%, which was the percentage of adults who used social media five years ago. Of the 2335 people who responded to survey, 1725 stated that they currently use social media.a) Use the data from this survey to construct a 93% confidence interval estimate of the proportion of adults who use social media. Record the result below in the form of (#,#)(#,#). Round your final answer to four decimal places.
- You recently sent out a survey to determine if the percentage of adults who use social media has changed from 62%, which was the percentage of adults who used social media five years ago. Of the 2283 people who responded to survey, 1635 stated that they currently use social media.a) Use the data from this survey to construct a 94% confidence interval estimate of the proportion of adults who use social media. Record the result below in the form of (#,#)(#,#). Round your final answer to four decimal places.If the number of smokers in a sample of 40 students selected from university A is 14 and the number of smokers in a sample of 380 students selected from university B is 84. Then the 99% confidence interval for the difference between proportions of smokers in the two universities (PB-PA) is closest to O a. (-0.3110,0.0534) O b. (-0.3310,0.0729) O c. (-0.1430,-0.1150) O d. (-0.1450,-0.1130) Oe. None of theseA professor was curious about her students' grade point averages. She took a random sample of 15 students and found a mean GPA of 3.01 with a standard deviation of 0.534. Which of the following formulas gives a 99% confidence interval for the mean GPA of the professor's students?
- A random sample of high school seniors were asked whether they were applying for college. The resulting confidence interval for the proportion of students applying for college is (0.65,0.69). What is the margin of error?In a survey of 2490 golfers, 22% said they were left-handed. The survey's margin of error was 4%. Construct a confidence interval for the proportion of left-handed golfers. O (0.17, 0.27) (0.18, 0.26) ○ (0.18, 0.22) ○ (0.26, 0.30)A random sample of 75 customers was polled. .626 and the 95% confidence interval