Question 150449
3.3048 is the sample mean= x with line over it.
1.32 is the standard deviation
1.833 is the “t” value**
10 is the sample size 

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 Page 309 
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.
A confidence interval looks like (x-bar-E < u < x-bar+E)
E is the margin of error and equals t * [s/sqrt(n)]
Your E = 1.833[1.32/sqrt(10)] = 0.76513...
Your 90% CI is (3.3048-0.76513 < u < 3.3048+0.76513)
This can be simplified: (2.5397 < u < 4.06993)
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(b) What sample size would be necessary to estimate the true weight with an error of ± 0.03 grams with 90 percent confidence? 
Using the formula for E you can solve for "n":
E = t * [s/sqrt(n)]
sqrt(n) = t*s/E
n = [ts/E]^2
Your problem:
n = [1.833*1.32/0.03] = 6504 when rounded down
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(c) Discuss the factors which might cause variation in the weight of Tootsie Rolls during manufacture. 
machinery error, weighing variation, mix inconsistency, etc.
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Cheers,
Stan H.