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000 AT&T; LTE 1:25 PM 99% mathxl.com Wei Shop Youn. Welcome t... Ims.ssc.ed... McGraw-H Ims.ssc.ed Homework Do Hom MTH 126-301, Fall 2017 Lisa Koscielski 11/2/17 1:25 PM Homework: Section 7.4 Homework Save Score: 8.57 of 12 pts 4 of 6 (6 complete) Hw Score: 71.9%, 21.57 of 30 pts 7.4.53 Question Help * A poll reported that only 680 out of a total of 1859 adults in a particular region said they had a "great deal of confidence" or "quite a lot of confidence" in the public School system This was down 5 percentage points from the previous year, Assume the conditions for using the CLT are met Complete parts (a) trough (4) below a. Find à 95% confidence interval for the proportion that express a great deal of confdence or quito a lot of confidence in the public schools, and interpret this The 95% confidence intervalfor the proporton that express agreat deal of confidence or qute alot of confidence inthe pubic schools is (0.344 0.388 Round to ree decimal places as needed) Interpret this interval. Sellect the correct choice below and ll in the answer bowes to complete your choice Type integers or decimals rounded to three decimal places as needed.) OA. %chance that the population proportion of adults having a goat deal or qite a lotofconfdence inghe public schools is between and . There is a We are 96 % confident that the population proportion of adults hav ng a great deal or quite a lot of confidence in thie public schools is between 0.344 and 0.388 There is a %chance that the sample proportion of aduls having a great deal or quite a lot ofconfidence inthe public schools is between and 11. we are %confident that the every sample proportion of adults having a great deal or quite a lot of confidence in the public schools is between and B OC. D. b. Find an 80% confidence interval and interpret The 80% confidence interval for the proporton that express a great deal of confidence or quite alot of confidence inthe pubic schools is (0351 0380 Round to tree decimal places as needed.) Interpret this interval Select the correct choice below and fill in the answer boxes to complete your choice. Type integers or decimals rounded to three decimal places as needed) OA We are %confident tha, tho overy sample proportion of adults having a great deal or quite a lot of confidence in the public schools is between and OB OC. D %chance that the population proportion of adults having a great deal or q te a lot of confdence in the public schools between and Tere is a There is a we are 80%confident that the population proportion ofad its hav ng a great deal or quite a lotofconfidence inthe public schools s between 0351 %chance that the sample proportion of adults having a great deal or qute a lot ofconfidence inthe public schools is between and and 0.380 C. Find the width of each interval by subtracting the lower proportion from the upper proportion, and state which interval is wider The widh of the 95% confdence interais 0 044 and the wdt, of the 80% confidence interal i5 0029 The 95% interval is wider Round to fhree decimal places as needed.) d, How would a 90% interval compare with the others in width? OA A90% interval wold be wider than the 95% confidence interval butnarrower than the 80% confidence because intervals get narrower with increasing B A9%interval would be narrower than the 95% confidence interval but wider than the 80% confidence because intervals get wider with increasing A90% interval would be narrower than the 95% confidence interval but wider than the 80% confidence because intervals get narrower with increasing A 90% interval would be wider than the 95% confidence interval but narrower than the 80% confidence because intervals get wider with increasing OC. O D Question is complete. Tap on the red indicators to see incorrect answersExplanation / Answer
p = 680/1859 = 0.37
P = 0.37 + 0.05 = 0.42
At 95% cinfidence interval the critical value Z* = 1.96
Cinfidence interval = P +/- Z* * sqrt (P * (1 - P )/n)
= 0.42 +/- 1.96 * sqrt(0.42 * 0.58 / 1859)
= 0.42 +/- 0.022
= 0.398, 0.442
At 80% confidence interval the critical value = 1.282
Cinfidence interval = P +/- Z* * sqrt (P * (1 - P )/n)
= 0.42 +/- 1.282 * sqrt(0.42 * 0.58 / 1859)
= 0.42 +/- 0.015
= 0.405, 0.435
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