Module 1 Assignment
Show all work. Use Minitab as much as you can.
(Montgomery & Runger, 2007)
Jaylen Rhodes
My Answers are in bold.
Question 1:
Eight measurements were made on the inside diameter of forged piston rings used
in an automobile engine. The data (in millimeters) are 74.001, 74.003, 74.015,
74.000, 74.005, 74.002, 74.005, and 74.004.
(a) Compute the sample mean and sample standard deviation of the temperature data.
x_bar = (74.001 + 74.003 + 74.015 + 74.000 + 74.005 + 74.002 + 74.005 +
74.004) / 8 = 592.035 / 8 = 74.004 mm
s^2 = ((74.001-74.004)^2 + (74.003-74.004)^2 + (74.015-74.004)^2 +
(74.000-74.004)^2 + (74.005-74.004)^2 + (74.002-74.004)^2 + (74.00574.004)^2 + (74.004-74.004)^2) / 7 = 0.000153 / 7 = 0.00002186
s = sqrt(s^2) = sqrt(0.00002186) = 0.00465794
(b) Graph a histogram of the data.
(c) Graph a normal probability plot of the data.
(d) Comment on the data.
Based on the probability plot above it seems most of the measured
values have a low variation between them and they all are relatively
close to the mean values. This seems to show a bit of consistency in
regards to manufacturing.
Question 2:
The April 22, 1991 issue of Aviation Week and Space Technology reports that during
Operation Desert Storm, U.S. Airforce F-117A pilots flew 1270 combat sorties for a
total of 6905 hours. What is the mean duration of an F-117A mission during this
operation? Why is the parameter you have calculated a population mean?
Mean duration = 6905 / 1270 = 5.437 hours
This value is a population mean because sample mean is just a reasonable
estimate of the population mean and we can see that based on the context of
the problem, that the 1270 combat sorties is the full population.
Question 3:
The yield of a chemical process is being studied. From previous experience yield is
known to be normally distributed and σ = 3. The past five days of plant operation
have resulted in the following percent yields: 91.6, 88.75, 90.8, 89.95, and 91.3.
Find a 95% two-sided confidence interval on the true mean yield.
x_bar = 90.48 from minitab, s = 1.15 from minitab, z_0.025 = 1.96 from z-table
95% CI:
x_bar ± z_0.025 * (σ/sqrt(n)) = 90.48 ± 1.96 * 3/sqrt(5) = 90.48 ± 1.96 * (1.34) =
90.48 ± 2.63 = (87.85,93.11)
Question 4:
A manufacturer produces crankshafts for an automobile engine. The wear of the
crankshaft after 100,000 miles (0.0001 inch) is of interest because it is likely to
have an impact on warranty claims. A random sample of n = 15 shafts is tested and
the mean equals = 2.78. It is known that σ = 0.9 and that wear is normally
distributed.
(a) Test H0 : μ = 3 versus H1: μ ≠ 3 using α = 0.05.
Null and Alternate Hypothesis:
H_0: μ = 3
H_1: μ ≠ 3
Descriptive Statistics
N
95% CI for μ
Mean SE Mean
15 2.780
0.232 (2.325, 3.235)
μ: population mean of Sample
Known standard deviation = 0.9
Test
Null hypothesis
H₀: μ = 3
Alternative hypothesis H₁: μ ≠ 3
Z-Value P-Value
-0.95
0.344
P-value is greater than the level of significance α = 0.05 so we do not reject the
null hypothesis H_0 which means it can be concluded that, there is no
significant difference in the means impact on the warranty claims.
(b) What is the power of this test if μ = 3.25?
Results
Difference Sample Size
0.25
Power
15 0.189511
The power of this test if μ = 3.25 is 0.189511.
Question 5:
The sodium content of thirty 300-gram boxes of organic corn flakes was
determined. The data (in milligrams) are as follows: 131.15, 130.69, 130.91,
129.54, 129.64, 128.77, 130.72, 128.33, 128.24, 129.65, 130.14, 129.29, 128.71,
129.00, 129.39, 130.42, 129.53, 130.12, 129.78, 130.92, 131.15, 130.69, 130.91,
129.54, 129.64, 128.77, 130.72, 128.33, 128.24, and 129.65.
(a) Can you support a claim that mean sodium content of this brand of
cornflakes differs from 130 milligrams? Use α = 0.05. Find the p-value.
x_bar = 129.75 from minitab, s = 0.9294 from minitab, n = 30
Null and Alternate Hypothesis
H_0: μ = 130
H_1: μ ≠ 130
Descriptive Statistics
N
Mean StDev SE Mean
30 129.753 0.929
95% CI for μ
0.170 (129.406, 130.100)
μ: population mean of C4
Test
Null hypothesis
H₀: μ = 130
Alternative hypothesis H₁: μ ≠ 130
T-Value P-Value
-1.46
0.156
Based on the test, the p-value is 0.156.
Since the p-value is greater than the level of significance α = 0.05, we do not
reject the null hypothesis H_0 and conclude that the mean sodium content of
this brand of cornflakes is not significantly different from 130 milligrams.
(b) Check that sodium content is normally distributed.
Based on this probability plot for the data, it does seem this data is normally
distributed.
(c) Compute the power of the test if the true mean sodium content is 130.5
milligrams.
Results
Difference Sample Size
0.5
Power
30 0.812694
The power of this test if the true mean sodium content is 130.5 milligrams is
0.812694.
Question 6:
A random sample of 300 circuits generated 13 defectives.
(a) Use the data to test H0: p = 0.05 versus H1: p ≠ 0.05. Use α = 0.05. Find the pvalue for the test.
Given n = 300, p_hat = 13 / 300 = 0.043
Null and Alternate Hypothesis:
H_0: p =0.05
H_1: p ≠ 0.05
I’m a bit confused on how to do this with minitab compared to the others so I
will do this by hand using the normal equation for test statistic and p-value
Test statistic:
z = (p_hat – p) / sqrt((p * (1-p))/ n)
z = (0.043 – 0.05) / sqrt((0.05 * 0.95) / 300) = -0.007 / sqrt(0.0475 /
300)
z = -0.007 / 0.013 = -0.556
P-value:
p_value = 2*P(Z>|z|) = 2(1-P(Z <0.556)
p_value = 2 * (1-0.702) = 0.596
Since the p-value is greater than the level of significance α = 0.05, we do
not reject the null hypothesis H_0 of p = 0.05.
(b) Explain how the question in part (a) could be answered with a confidence
interval.
We can calculate a two-sided confidence interval and if the confidence
interval includes the value 0, we do not reject H_0 but if it does not then
we do reject H_0.
Question 7:
Consider the hypothesis test H0: μ1 = μ2 against the H1: μ1 ≠ μ2. Suppose that sample
sizes are n1 = 15 and n2 = 15, that the mean of 1 = 4.7 and the mean of 2 = 7.8 and
that s12 = 4 and s22 = 6.25. Assume that σ12 = σ22 and that the data are drawn from
normal distributions. Use α = 0.05.
(a) Test the hypothesis and the find the p-value.
Null and Alternate Hypothesis:
H_0: μ1 = μ2
H_1: μ1 ≠ μ2
s^2_p = (n_1 – 1)*s^2_1 + (n_2 – 1)*s^2_2 / (n_1 + n_2 – 2)
s^2_p = (15-1)*4 +(15-1)*6.25 / (15 + 15 – 2)
s^2_p = (56 + 87.5) / 28 = 5.125
s_p = sqrt(5.125) = 2.264
Test Statistic:
t_0= (x_bar_1 - x_bar_2) / (s_p*sqrt(1/n_1 + 1/n_2))
t_0= (4.7 – 7.8) / (2.264 * sqrt(1/15 + 1/15))
t_0 = -3.1 / 0.827 = -3.748
P-value:
p_value = 2*P(t <= -3.748) = 2 (0.00009)
p_value = 0.00018
Since the p-value is less than the level of significance α = 0.05, we reject
the null hypothesis H_0 of μ1 = μ2 and instead accept the alternate
hypothesis H_1 of μ1 ≠ μ2.
(b) Explain how the test could be conducted with a confidence interval.
Similar to the last problem, we can calculate a two-sided confidence interval
and if the interval includes 0 we accept H_0, but if it does not we would reject
H_0.