Question 150449: I found where someone answered this problem...Stan H. However, I didn't understand how he got the answer. I really would like to understand the how to on this since this is a statics class and I have more to do. On the random sample question I know that I needed to get the sample mean..standard deviation, t value and sample size. I have posted it below. Then I have to use the forumla x ± t s√n....(that is suppose to be a square root sign with the s over the n.) X is the sample mean...and t is the t value...but then I'm not sure where to place the standard deviation or the sample size... are they the s or the n...then which part of the equation do you do first and what does this little sign mean ±? This part is only for section A. I have no idea how to work through section b. And yes, I understand c with no problem. :) I really appreciate in advance any help!
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.
(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 stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 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.
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