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Question 153491: A random sample of 10 miniature Tootsie Rolls was taken from a bag. Each piece was weighed on a very accurate scale. The results in grams were
3.087 3.131 3.241 3.241 3.270 3.353 3.400 3.411 3.437 3.477
a). Construct a 90 percent confidence interval for the true mean weight. (b) What sample size would be necessary to estimate the true weight with an error of ± 0.03 grams with 90 percent confidence? (c) Discuss the factors which might cause variation in the weight of Tootsie Rolls during manufacture.: A random sample of 10 miniature Tootsie Rolls was taken from a bag. Each piece was weighed on a very accurate scale. The results in grams were
3.087 3.131 3.241 3.241 3.270 3.353 3.400 3.411 3.437 3.477
a). Construct a 90 percent confidence interval for the true mean weight. (b) What sample size would be necessary to estimate the true weight with an error of ± 0.03 grams with 90 percent confidence? (c) Discuss the factors which might cause variation in the weight of Tootsie Rolls during manufacture.
Answer by Fombitz(1755) About Me  (Show Source):
You can put this solution on YOUR website!
a)Since you don't know sigma and your sample size is <30, assume your samples are normally distributed.
Use the t-distribution as the best estimate. Degrees of freedom are n-1 or 9.
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The calculated mean,x[m], is 3.3048.
The calculated standard deviation, s, is 0.132
For alpha=0.10 and DOF=9, t=1.833.
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The confidence interval is
x[m]-t(s/sqrt(n))< mu < x[m]+t(s/sqrt(n))
3.3048-1.833(0.132/sqrt(10))< mu < 3.3048+1.833(0.132/sqrt(10))
3.228<mu <3.381
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b) Working backwards,
t(s/sqrt(n))=0.03
t(0.132/sqrt(n))=0.03
t/sqrt(n)= 0.227
t is a function of n so you have to iterate to find n.
I set up an iteration cell in EXCEL using TINV and varying n.
n=54 gives t/sqrt(n)= 0.227743
54 samples required to give 0 +- 0.03
For this large a value for n, we can use the normal distribution as a check of the value since as n gets large the t distribution approaches the normal distribution.
n=((z*(sigma))/0.03)^2
z=1.65 for 90%
Use the calculated 0.132 as an estimate for sigma
n=((1.65*(0.132))/0.03)^2
n=(7.26)^2
n=52.7 or n=53
Good, that's close.
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c) Factors include length errors from cutting machine, air in the mixture, humidity in the plant, density changes in the mixture, etc.