SOLUTION: # 1.) The weight of full boxes of frosted Mini-wheat cereal are normally distributed with a standard deviation of 0.52 oz. A sample of 18 randomly selected boxes produced a mean we

Algebra ->  Probability-and-statistics -> SOLUTION: # 1.) The weight of full boxes of frosted Mini-wheat cereal are normally distributed with a standard deviation of 0.52 oz. A sample of 18 randomly selected boxes produced a mean we      Log On


   

Question 314396: # 1.) The weight of full boxes of frosted Mini-wheat cereal are normally distributed with a standard deviation of 0.52 oz. A sample of 18 randomly selected boxes produced a mean weight of 24.3 oz. Find the 95% confidence interval for the true mean weight of a box of this cereal.
# 2.) We are interested in estimating the mean life of a new product. How large a sample do we need to take in order to estimate the mean to within 0.20 of a standard deviation with 90% confidence?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
# 1.) The weight of full boxes of frosted Mini-wheat cereal are normally distributed with a standard deviation of 0.52 oz. A sample of 18 randomly selected boxes produced a mean weight of 24.3 oz. Find the 95% confidence interval for the true mean weight of a box of this cereal.
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sample mean: 24.3
E = [invT(0.975 with df = 17)]*0.52/sqrt(18) = 2.1098*0.1226 = 0.2586
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95% CI: 24.3-0.2596 < u < 24.3+0.2596
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# 2.) We are interested in estimating the mean life of a new product. How large a sample do we need to take in order to estimate the mean to within 0.20 of a standard deviation with 90% confidence?
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n = [z*s/E]^2
n = [1.645*s/0.2]^2
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Note: You need to know the standard deviation.
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Cheers,
Stan H.