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a. If n is even, show that (3^n-1)2 is always divisible by 4, so it.pdf
a. If n is even, show that (3^n-1)2 is always divisible by 4, so it.pdfCargando en ... 3
a. determine the sample proportionb. decide whether using the one .pdf
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a. determine the sample proportion b. decide whether using the one proportion z interval procedure is apporapriate c. If appropriate, use the one proportion z interval procedure to find the confidence interval at the specified confidence level. x=40, n=50, 95% level x=3, n=100, 99% level show work Solution (a) p=40/50=0.8 p=3/100=0.03 (b)one proportion z interval procedure is apporapriate (c) Given a=0.05, |Z(0.025)|=1.96 (From standard normal table) So 95% CI is p +/- Z*v(p*(1-p)/n) --> 0.8+/- 1.96*sqrt(0.8*0.2/50) --> (0.6891257, 0.9108743) Given a=0.01, |Z(0.005)|=2.58 (from standard normal table) So 99% CI is p +/- Z*v(p*(1-p)/n) --> 0.8+/- 2.58*sqrt(0.8*0.2/50) --> (0.6540532, 0.9459468).
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a. determine the sample proportionb. decide whether using the one .pdf
- a. determine the sample proportion b. decide whether using the one proportion z interval procedure is apporapriate c. If appropriate, use the one proportion z interval procedure to find the confidence interval at the specified confidence level. x=40, n=50, 95% level x=3, n=100, 99% level show work Solution (a) p=40/50=0.8 p=3/100=0.03 (b)one proportion z interval procedure is apporapriate (c) Given a=0.05, |Z(0.025)|=1.96 (From standard normal table) So 95% CI is p +/- Z*v(p*(1-p)/n) --> 0.8+/- 1.96*sqrt(0.8*0.2/50) --> (0.6891257, 0.9108743) Given a=0.01, |Z(0.005)|=2.58 (from standard normal table) So 99% CI is p +/- Z*v(p*(1-p)/n) --> 0.8+/- 2.58*sqrt(0.8*0.2/50) --> (0.6540532, 0.9459468)
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