SOLUTION: In a sample of 100 Planter’s Mixed Nuts, 19 were found to be almonds. (a) Construct a 90 percent confidence interval for the true proportion of almonds. (b) May normality be assume

Algebra ->  Probability-and-statistics -> SOLUTION: In a sample of 100 Planter’s Mixed Nuts, 19 were found to be almonds. (a) Construct a 90 percent confidence interval for the true proportion of almonds. (b) May normality be assume      Log On


   

Question 304872: In a sample of 100 Planter’s Mixed Nuts, 19 were found to be almonds. (a) Construct a 90 percent confidence interval for the true proportion of almonds. (b) May normality be assumed? Explain.
(c) What sample size would be needed for 90 percent confidence and an error of ± 0.03? (d) Why would a quality control manager at Planter’s need to understand sampling?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
In a sample of 100 Planter’s Mixed Nuts, 19 were found to be almonds.
(a) Construct a 90 percent confidence interval for the true proportion of almonds.
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p-hat = 0.19
standard error = 1.645*sqrt(0.19*0.81/100) = 0.0645
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90% CI: 0.19-0.0645 < p < 0.19+0.0645
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(b) May normality be assumed? Explain.
I'll leave that to you.
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(c) What sample size would be needed for 90 percent confidence and an error of ± 0.03?
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n = [z/E]^2*pq
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n = [1.645/0.03]^2*0.19*0.81 = 462.65
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Round up to n = 463
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(d) Why would a quality control manager at Planter’s need to understand sampling?
Testing will show him whether the product is as it is being advertised.
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Cheers,
stan H.
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