fAnd a salunan onyCUrawb memorization of1le salution indipevelmm> ploblem solving dipendem meihod hin M-10-50 M- 6 16 SS-108 SS-116 used a.is there a signifiant diffeichce in performance on new prabeims for these TWO groupse Use a two-tauled rest with a-.05 Null hypotnesas- Hoi lli-llz-0 Altematire hypanis>Hii lli-elz#0 poolld variance→ S*P = SSi+SSz = 224- 16 14 %3D 108+ 116 %3D afi tdfz 717 estimatedstandara error» S(MI - M2) =Sp+s?P 16 16 32-4=2 ni -Stats→ t= (MI-M2)+i-Uz) S(MI-M2) 10.5-6.16 4.34 -2.17 find the chraevalue cf=8+8-2%3D144 20145-criticaevalue The 1-stat vasue is mgher than the criticaevalue, turciore wemust reject ihe hull nypothls1s. In conclusion. thereis a sighifice nt dif Ferchce in perfarmanCI on new problems for thigirengroups aconstruct a 907• Confidlence inteIval to cs Timate the s1ze Of the meandifference CIl,-llz=MI-M21ts (MI-M2) 0 = 10:5-6.16 2.17 (2) 나.3나 4.3나 (8.681 0)

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Construct a 96% confidence interval if x=75 and n=200. Round your answers to 3 decimal places.

A poll of 1070 U.S. adults found that 563 U.S. adults, or 52.617% regularly used Facebook as a news source. Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 90% level of confidence. Express all answers as percentages rounded to 2 decimal places. Do not write the percent sign. Margin of Error (as a percentage): Confidence Interval: Margin of Error (as a percentage): Confidence Interval: Now find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 95% level of confidence. Margin of Error (as a percentage): Confidence Interval: % to O increases O stays roughly the same O decreases % to Finally, find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 99% level of confidence. and the precision of our estimate O decreases O increases O stays roughly the…