For this problem, assume alpha = 0.1 (i.e assume 90% significance) A machine sho

Question

For this problem, assume alpha = 0.1 (i.e assume 90% significance) A machine shop is required to produce shafts with diameter 25 mm. A sample of 9 parts is randomly selected from a batch of parts. The diameters of the parts are measured as follows: 24.84, 24.86, 24.89, 24.93, 24.95, 24.98, 25.02, 25.04, 25.06. a. If the standard deviation of the lathe is known to be 0.1 mm, determine the 90% confidence interval of the mean? b. If the standard deviation of the lathe is not known, determine the 90% confidence interval of the mean? c. If the standard deviation of the lathe is known to be 0.1 mm, determine the probability that the sample mean is between 24.95 and 25.05 d. If the standard deviation of the lathe is unknown, determine the probability that the sample mean is between 24.95 and 25.05. e. What is the probability that a given part's diameter exceeds 25.12 mm if the standard deviation is 0.1 mm? f. What is the probability that a given part's diameter is between 24.85mm and 24.95 mm if the standard deviation is 0.1 mm?

Explanation / Answer

x

24.84

24.86

24.89

24.93

24.95

24.98

25.02

25.04

25.06

Total =

224.57

X bar = 224.57 / 9 = 24.9522

Question a)

Here standard deviation is known so we will use z-distribution to find the confidence interval.

X bar (-/+) E

E = zc * ( sigma / sqrt (n))

From normal table we get the critical z for 10% level of significance as 1.645

E = 1.645 * (0.1/sqrt(9)) = 0.0548

X bar (-/+) E

24.9522 (-/+) 0.0548

24.897 and 25.007

Answer: 24.897 and 25.007

Question b)

Here sample standard deviation is unknown so we will use t-distribution to find the confidence interval.

We use excel to find the sample standard deviation;

=stdev(data set)

s = 0.07918

X bar (-/+) E

E = tc * ( s / sqrt (n))

We find the critical t for 10% level of significance for 8 degrees of freedom as 1.860

E = 1.860*( 0.07918/sqrt(9)) = 0.0491

X bar (-/+) E

24.9522 (-/+) 0.0491

24.903 and 25.001

Answer: 24.903 and 25.001

Question c)

P ( 24.95 < x bar < 25.05)

z1 = ( x bar – Mean)/(sigma/sqrt(n))

  = (24.95 – 24.9522)/(0.1/sqrt(9))

    =-0.07

z2 = ( x bar – Mean)/(sigma/sqrt(n))

    = (25.05 – 24.9522)/(0.1/sqrt(9))

    =2.93

P (-0.07 < z < 2.93 )

= P ( z < 2.93 ) – P ( z < -0.07)

= 0.9983 – 0.4721                    [ By using Normal Table]

= 0.5262

Answer: 0.5262

Question d)

P ( 24.95 < x bar < 25.05)

z1 = ( x bar – Mean)/(s/sqrt(n))

    = (24.95 – 24.9522)/(0.07918/sqrt(9))

    =-0.08

z2 = ( x bar – Mean)/(s/sqrt(n))

    = (25.05 – 24.9522)/(0.07918/sqrt(9))

    =3.71

P (-0.08 < z < 3.71 )

= P ( z < 3.71 ) – P ( z < -0.08)

= 0.9999 – 0.4681                    [ By using Normal Table]

= 0.5318

Answer: 0.5318

x

24.84

24.86

24.89

24.93

24.95

24.98

25.02

25.04

25.06

Total =

224.57

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