Question 1065358
1. One sample proportion, p=0.45, 1-p=0.55
interval width for 95% is z(0.975)* sqrt ((p*(1-p))/n)=1.96* sqrt (0.2475/400)=0.0249*1.96=0.0488
(0.4012, 0.4988).
For 99%, z=2.576 and interval width is 0.0641 for (0.3859, 0.5141)
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2. The 95% CI is x bar +/- t (df=24, 0.975)*2.01/sqrt (25)
interval width is 2.064*2.01/5=0.83
95% interval is (4.07, 5.73)
90% interval uses 1.711 for t and has interval width of 0.70; (4.2, 5.6)
99% interval uses 2.797 for t and has interval width of 1.12; (3.78, 6.02)
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3. This doesn't appear to work with mean of 4000 and sd of 63,245; if such an sd is in fact the case, the distribution is not normal and can't use that. I think the 63.245 would likely be too small a sd.
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4.Mean is 1.75 m and sigma is 0.4 m (sqrt of 0.16 m^2)
95% CI interval width is is 1.96*0.4/20=0.0392
(1.71, 1.79)
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Need an interval width of 2 with z=1.645, sd 0.4 and n unknown
2=1.645*0.4/ sqrt (n)
sqrt (n)=1.645*0.4/0.02=32.9
square both sides, and n=1082.41=1083.