Using the t table Find the degrees of freedom (df) and the value of t* for the g
ID: 3221117 • Letter: U
Question
Using the t table Find the degrees of freedom (df) and the value of t* for the given sample size and confidence level or significance level (alpha). (a) n = 6,CL = 90% (b) n = 21,CL = 98% (c) n = 29, CL = 95% (d) n = 12, CL = 99% (e) n = 6, alpha = 0.10 (f) n = 21, alpha = 0.01 (g) n = 40, alpha = 0.05 Confidence Interval A dealer in recycled paper places empty trailers at various sites. The trailers are gradually filled by individuals who bring in old newspapers and magazines, and are picked up on several schedules. One such schedule involves pickup every second week. This schedule is desirable if the average amount of recycled paper is more than 1600 cubic feet per 2-week period. The dealer's records for eighteen 2-week periods show the following volumes (in cubic feet) at a particular site: recycle = c (1660,1820,1590,1440,1730, 1680, 1750, 1720, 1900, 1570, 1700, 1900, 1800, 1770, 2010, 1580, 1620, 1690) The mean and standard deviation are as follows X = 1718.3 and s = 137.8 (a) In constructing confidence intervals, would we use 2 or t in this situation? Briefly explain why you would use one instead of the other. (b) Estimate the true mean weight of recycled paper with 95% confidence. Interpret.Explanation / Answer
4)
a) n = 6 , Cl =90%
degrees of freedom = n -1 = 6 -1 = 5
Value of t using df = 5 , CL = 90% = +/-2.015
b) n = 21 , CL =98%
degrees of freedom = n -1 = 21 -1 = 20
Value of t using df = 20 , CL = 98% = +/- 2.528
c)
n = 29 , CL =95%
degrees of freedom = n -1 = 29 -1 = 28
Value of t using df = 28 , CL = 95% = +/- 2.048
d)
n = 12 , CL =99%
degrees of freedom = n -1 = 12 -1 = 11
Value of t using df = 11 , CL = 99% = +/- 3..106
e)
n = 6 , alpha = 0.10
degrees of freedom = n -1 = 6 -1 =5
Value of t using df = 5 , alpha = 0.10 = +/- 2.015
f)
n = 21, alpha = 0.01
degrees of freedom = n -1 = 21 -1 = 20
Value of t using df = 20 , alpha = 0.01 = +/- 2.845
g)
n = 40, alpha = 0.05
degrees of freedom = n -1 = 40 -1 =39
Value of t using df = 39 , alpha = 0.05 = +/- 2.021
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