Question 312937: 1. By measuring the amount of time it takes a component of a product to move from one workstation to the next, an engineer has estimated that the standard deviation is 1.8 seconds.
(a)How many measurements should be made in order t be 99% certain that the maximum error of estimation will not exceed 0.5 seconds?
(b) What sample size is required for a maximum error of 1 second?
2. In testing the hypothesis H0: u _>35 and Ha: u <35 using the p-value approach, a p - value of 0.0721 was obtained. If standard deviation = 8.1 find a sample mean which produced this p - value given that the sample of size n = 32 was randomly selected.
3.A study was conducted t estimated the mean amount spent on birthday gifts for a typical family having two children. A sample of 125 was taken, and the mean amount spent was $230. Assuming a standard deviation equal to $45 find the 95% confidence interval for u the mean for all such families.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1. By measuring the amount of time it takes a component of a product to move from one workstation to the next, an engineer has estimated that the standard deviation is 1.8 seconds.
(a)How many measurements should be made in order to be 99% certain that the maximum error of estimation will not exceed 0.5 seconds?
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n = [z*s/E]^2
n = [2.5758*1.8/0.5]^2 = 86 when rounded up
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(b) What sample size is required for a maximum error of 1 second?
Comment: Same procedure but use "1" in place of 0.5
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2. In testing the hypothesis H0: u >= 35 and Ha: u <35 using the p-value approach, a p-value of 0.0721 was obtained. If standard deviation = 8.1 find a sample mean which produced this p-value given that the sample of size n = 32 was randomly selected.
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Find the t-value with a left tail of 0.0721 when df = 31 : t = -1.4982
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Find the sample mean:
(x-bar-35)/[8.1/sqrt(31)] = -1.4982
x-bar-35 = -2.1796
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x-bar = 32.8204
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3.A study was conducted to estimate the mean amount spent on birthday gifts for a typical family having two children. A sample of 125 was taken, and the mean amount spent was $230. Assuming a standard deviation equal to $45 find the 95% confidence interval for u the mean for all such families.
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sample mean: 230
E = (invT(0.975,124)*45/sqrt(125)) = 7.9664
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95% CI:
230-7.9664 < u < 230+7.9664
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Cheers,
Stan H.
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