Measurements of a critical dimension on a sample of automotive parts are taken w
ID: 3240754 • Letter: M
Question
Measurements of a critical dimension on a sample of automotive parts are taken with the following results (in units of mm):
10.14 10.4 9.87 10.02 9.99 10.03 10.09 10.11 9.93 9.99 10 10.02
The parts are produced from a machine known to have a standard deviation of = 0.10 mm. The intended is 10, but is known to drift from this mean during production. Based on this information, answer the following questions, clearly showing your work:
a. Calculate the 95% confidence interval for the mean for the production process during the time when these parts were machined.
b. Assume that the standard deviation for the production process is not known. Calculate the 95% confidence interval for the mean.
c. Assume that the standard deviation for the production process is known. Calculate the 99% confidence interval for the mean.
d. Assume that the standard deviation for the production process is not known. Calculate the 99% confidence interval for the mean.
e. Find the p-value (the smallest level of significance at which H0 would be rejected) under the assumption that is known and when is not known
f. 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.
Explanation / Answer
(a)
n = 12
x-bar = 10.05
s = 0.1
% = 95
Standard Error, SE = /n = 0.1 /12 = 0.028867513
z- score = 1.959963985
Width of the confidence interval = z * SE = 1.95996398454005 * 0.0288675134594813 = 0.056579287
Lower Limit of the confidence interval = x-bar - width = 10.05 - 0.0565792867038086 = 9.993420713
Upper Limit of the confidence interval = x-bar + width = 10.05 + 0.0565792867038086 = 10.10657929
(b)
n = 12
x-bar = 10.05
s = 0.133
% = 95
Standard Error, SE = s/n = 0.133/12 = 0.038393793
Degrees of freedom = n - 1 = 12 -1 = 11
t- score = 2.200985159
Width of the confidence interval = t * SE = 2.20098515872184 * 0.0383937929011101 = 0.084504168
Lower Limit of the confidence interval = x-bar - width = 10.05 - 0.0845041683623834 = 9.965495832
Upper Limit of the confidence interval = x-bar + width = 10.05 + 0.0845041683623834 = 10.13450417
(c)
n = 12
x-bar = 10.05
s = 0.1
% = 99
Standard Error, SE = /n = 0.1 /12 = 0.028867513
z- score = 2.575829304
Width of the confidence interval = z * SE = 2.57582930354892 * 0.0288675134594813 = 0.074357787
Lower Limit of the confidence interval = x-bar - width = 10.05 - 0.0743577870895246 = 9.975642213
Upper Limit of the confidence interval = x-bar + width = 10.05 + 0.0743577870895246 = 10.12435779
(d)
n = 12
x-bar = 10.05
s = 0.133
% = 99
Standard Error, SE = s/n = 0.133/12 = 0.038393793
Degrees of freedom = n - 1 = 12 -1 = 11
t- score = 3.105806514
Width of the confidence interval = t * SE = 3.10580651358217 * 0.0383937929011101 = 0.119243692
Lower Limit of the confidence interval = x-bar - width = 10.05 - 0.119243692073393 = 9.930756308
Upper Limit of the confidence interval = x-bar + width = 10.05 + 0.119243692073393 = 10.16924369.