please help! 2. (24 points, 4 each) Measurements of a critical dimension on a sa

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please help!

2. (24 points, 4 each) Measurements of a critical dimension on a sample of automotive parts are taken with the following results (in units of mm): 10.4 10.11 10.14 10.09 9.87 9.93 10.02 9.99 9.99 10 10.03 10.02 The parts are produced from a machine known to have a standard deviation of ? 0.10 mm. The intended u is 10, but is known to drift from this mean during production. Based on this information, answer the following questions, clearly showing your work: Calculate the 95% confidence interval for the mean for the production process during the time when these parts were machined. Assume that the standard deviation for the production process is not known. Calculate the 95% confidence interval for the mean Assume that the standard deviation for the production process is known. Calculate the 99% confidence interval for the mean Assume that the standard deviation for the production process is not known Calculate the 99% confidence interval for the mean Find the p-value (the smallest level of significance at which Ho would be rejected) under the assumption that ? is known and when ? is not known Find ? (the probability of a Type II error) for the case where a 95% confidence interval is used, the true mean 10.2, and ? is known b. d. e.

Explanation / Answer

a)
Two Sided Confidence Interval for 95%
Here, n = 12 , xbar = 10.0492 and s = 0.1
t-value is 2.201

Standard Error, SE = s/sqrt(n)
SE = 0.1/sqrt(12)
SE = 0.0289

Margin of Error, ME = t*SE
ME = 2.201*0.0289
ME = 0.0636

CI = (xbar - ME, xbar + ME)
CI = (10.0492 - 0.0636 , 10.0492 + 0.0636)
CI = (9.9856 , 10.1128)

b)
Two Sided Confidence Interval for 95%
Here, n = 12 , xbar = 10.0492 and s = 0.133
t-value is 2.201

Standard Error, SE = s/sqrt(n)
SE = 0.133/sqrt(12)
SE = 0.0384

Margin of Error, ME = t*SE
ME = 2.201*0.0384
ME = 0.0845

CI = (xbar - ME, xbar + ME)
CI = (10.0492 - 0.0845 , 10.0492 + 0.0845)
CI = (9.9647 , 10.1337)

c)
Two Sided Confidence Interval for 99%
Here, n = 12 , xbar = 10.0492 and s = 0.1
t-value is 3.1058

Standard Error, SE = s/sqrt(n)
SE = 0.1/sqrt(12)
SE = 0.0289

Margin of Error, ME = t*SE
ME = 3.1058*0.0289
ME = 0.0898

CI = (xbar - ME, xbar + ME)
CI = (10.0492 - 0.0898 , 10.0492 + 0.0898)
CI = (9.9594 , 10.139)

d)
Two Sided Confidence Interval for 99%
Here, n = 12 , xbar = 10.0492 and s = 0.133
t-value is 3.1058

Standard Error, SE = s/sqrt(n)
SE = 0.133/sqrt(12)
SE = 0.0384

Margin of Error, ME = t*SE
ME = 3.1058*0.0384
ME = 0.1193

CI = (xbar - ME, xbar + ME)
CI = (10.0492 - 0.1193 , 10.0492 + 0.1193)
CI = (9.9299 , 10.1685)

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