STAT360 Exam 2 Zo.os-1.645; Zo.025-1.96; Zo.o05-2.58 0.025-I .96 0.005-2.58 Tabl
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STAT360 Exam 2 Zo.os-1.645; Zo.025-1.96; Zo.o05-2.58 0.025-I .96 0.005-2.58 Table A.5 Critical Values for t Distributions density curve Shaded area = .10 .05 025 .01 005 001 0005 6.314 2.920 2.353 2.132 2.015 1.943 1.895 1.860 12.706 4.303 3.182 2.776 2.571 2.447 2.365 2.306 63.657 9.925 5.841 4.604 4.032 3.707 3.499 3.355 318.31 3.078 1.886 1638 1.533 1.476 1.440 1.415 1.397 31.821 6.965 4.541 3.747 3.365 3.143 2.998 2.896 636.62 22.326 10.213 7.173 5.893 5.208 4.785 4.501 31.598 12.924 8.610 6.869 5.959 5.408 5.041 Table A.8 t Curve Tail Areas (cont.) r curve Area to the right ofr 19 20 2 22 23 24 25 26 7 28 29 30 35 40 60 120) 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 461 461 461 46 46 46 461 461 461 461 46 46460 460 460 460 460 422 422 422 422 422 422 422 422 421 421 421 421 42 42 421 421 421 384 384 384 383 383 383 383 383 383 383 383 383 383 383 383 382 382 347 .347 .347 347 346 346 346 346 346 346 346 .346 346 346 345 345 .345 311 311 31 31 311 311 311 31 310 310 310 310 310 309 30 309 025 024 024 024 023 023 023 023 023 022 022 022 022 .021 020 019 018 .020 .020 .020 .019 .019 .019 .019 .018 .018 .018 .018 .018 .017 .017 .016 .015 .014 016 016 016 016 015 015 015 015 015 015 014 014 014 013 012 012 011 013 013 013 013 012 012 .012 012 012 .012 012 0 .011 011 010 009 008 011 011 010 010 010 010 010 010 009 009 009 009 009 008 008 007 006 0.0 0.1 0.2 0.5 2.2 2.3 2.4 2.5Explanation / Answer
for normal distribution' a score =(X-mean)/std deviaiton
5) probability =P(X<10)=P(Z<(10-10)/0.2)=P(Z<0) =0.5
6)P(9.75<X<10.15)=P((9.75-10)/0.2<Z<(10.15-10)/0.2)=P(-1.25<Z<0.75)=0.7734-0.1057 =0.6677
7)for 99th percentile z score =2.3263
therefore corresponding weight =mean +z*std deviaiton=10.465
8) expected weight of 4 cans =4*10=40
and std deviation =0.2*(4)1/2 =0.4
therefore probability =P(X>41)=1-P(X<41)=1-P(Z<(41-40)/0.4)=1-P(Z<2.5)=1-0.9938 =0.0062
9) parameter of interest is population proportion of satisfied customer of internet service provider.
10)
here estimated proportion phat=67/100=0.67
std error =(phat*(1-phat)/n)1/2 =0.047
for 90% CI ; critical z =1.6449
therefore 90% confidence interval =sample proporiton -/+ z*std error =0.5927 to 0.7473
11) above interval gives 90% confidence to contain true population proportion of satisfied customer of internet service provider.
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