I\'ve been on these few homework problems for over an hour and still can not fig

Question

I've been on these few homework problems for over an hour and still can not figure them out. Much help would be appreciated on how they can be solved. 1. What is the Critical Value of T that satisfies: 9 Degrees of Freedom, and probability of 0.10 to the right of T? 2.What is the minimum sample size needed to get a margin of error less than .0001, with an STD = 0.01, and mean of 1? 3.What is the standard error of the mean with N = 9, Mean = 165, and STD = 15 4.What is the minimum sample size needed to get a margin of error less than 10, with a 95 confidence interval, and and STD = 100 5.A survey is taken of whether or not a person uses the toilet three times a day or more, 60 people are asked, and 28 say yes. What is the Standard Error of the population (P.hat) 6. Construct a 96% Confidence interval in which 435 out of 1183 have a car. 7. Construct a 98% Confidence interval in which 448 out of 1280 have a car.

Explanation / Answer

1. Look at a table of T distribution criticl values or use a graphing calaculator invT(area, df) = invT(0.9, 9) Area represents to the left of T, df = degrees of freedom T = 1.383 . Standard error = s/sqrt(n) = 15/sqrt(9) = 5 z*s/sqrt(n) = m 1.96*100/sqrt(n) = 10 n = 384.16 round up so minimum sample size = 385 5. p^ = 28/60 = 0.467, n = 60 Standard error of proportion = sqrt( p^(1-p^)/n ) = sqrt(0.467*0.533/60) = 0.644 6. This is a one proportion z-test. p (proportion of population) is unknown. Interval is determined by p^ ± z*sqrt(p^(1-p^)/n) p^ = 435/1183 = 0.3677 To find z*, you can look at a z-table -- using area 1- 0.04/2 = 0.98 so z* = 2.054 7. Same strategy as #6.

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