
Question 1
10 out of 10 points
If the probability that any individual will react positively to a drug is 0.5, what is
the probability that 5 individuals will react positively from a sample of 10
individuals?
Answer
Selected Answer:
Correct Answer:
24.6%
[None]
Response Feedback: [None Given]

Question 2
10 out of 10 points
In a sample of 100 items produced by a machine that produces 10% defective
items, what is the probability that 8 items are defective? (Calculate with
binomial distribution formula and verify your response using MINITAB).
Attach your answer as a word document/pdf file. Appended at end
Answer
Selected Answer:
Question 2.docx
Response Feedback: [None Given]

Question 3
10 out of 10 points
Attach your answer as a Word document/PDF file. Appended at end
Answer
Selected Answer:
Question 3.docx
Response Feedback: [None Given]

Question 4
10 out of 10 points
The American Association of University Professors claims that the mean income
of tenured professors at public universities is $ 65,000. Our hypothesis is that
the mean salary is actually lower than $65,000. In order to test whether or not
the mean salary is lower than $ 65,000. We take a random sample of n=40
professors. Their salaries led to a sample average of $60,000 and a sample
standard deviation s=$8,000.
(a) Calculate the test statistic and obtain its probability value.
(b) Assuming a significance level of 5 percent, what is your conclusion?
Answer
Selected
Answer:
One-Sample T
Test of mu = 65000 vs not = 65000
N Mean StDev SE Mean 95% CI T P
40 60000 8000 1265 (57441, 62559) -3.95 0.000
Correct
With a 5% level of significance, we can conclude
that the alternative hypothesis (salary < 65,000)
is true
[None]
Answer:
Response
Feedback:

[None Given]
Question 5
9 out of 10 points
Thirty light bulbs were selected randomly form among a very large production
batch, and they were put on test to determine the time until they burn out. The
average failure time for these thirty bulbs was 1,100 hours; the sample standard
deviation was 200 hours.The light bulbs are advertised as having a mean life
length of 1150 hours. Test this hypothesis against the alternative that the mean
life length of this batch is actually smaller than 1,150 hours.
Answer
Selected Answer: Test of mu = 1150 vs < 1150
N Mean StDev SE Mean 95% Upper Bound
30 1100.0 200.0 36.5
1162.0
T
P
-1.37 0.091
with a 95% confidence level, the bulbs mean life is greater
than 1150.
Correct Answer: [None]
Response
Feedback:
Correct Ans:
P (z= - 1.37) = 0.085
0.085 > 0.05; therefore, we accept the null hypothesis

Question 6
10 out of 10 points
Question 6
Answer
Selected
Answer:
Test of mu = 15 vs < 15
N Mean StDev SE Mean 95% UpperBound
25 14.200 1.300 0.260
Correct
Answer:
Response
Feedback:

14.645
T
P
-3.08 0.003
With a 95% confidence interval, the mean of the
population is less than 15 ounces
[None]
[None Given]
Question 7
10 out of 10 points
Question no 7
Answer
Selected
Answer:
Test of mu = 22 vs > 22
N Mean StDev SE Mean 95% Lower Bound
T
P
15 25.80 4.70 1.21
Correct
Answer:
Response
Feedback:

23.66
3.13 0.004
With a 95% confidence interval, the average time
it will take is over 22.0 minutes.
[None]
[None Given]
Question 8
9 out of 10 points
Question no 8
Answer
Selected
Answer:
Test of mu = 2 vs > 2
N Mean StDev
SE Mean 95% Lower Bound
180 1.7000 0.3000 0.0224
1.6630
T
P
-13.42 1.000
With a 95% confidence interval, the calls are
less than 2 minutes
Correct Answer: [None]
Response
Feedback:
[None Given]

Question 9
10 out of 10 points
A diet plan states that, on average, participants will lose 13 pounds in four
weeks. As a statistician for a competing organization, you want to test the
claim. You sample 100 participants who have been on the subject diet plan for
four weeks. You find that the average weight loss has been 1 1.4 pounds with a
sample standard deviation of 1.6 pounds. What is your conclusion based on the
data using a 5 percent level of significance (risk)?
Answer
Selected
Answer:
Test of mu = 13 vs < 13
N Mean StDev SE Mean
100 11.400 1.600 0.160
Correct
Answer:
Response
Feedback:

Question 10
10 out of 10 points
Question no 10
95% Upper Bound
11.666
T
P
-10.00 0.000
With a 95% confidence interval, the average
weightloss is less than the advertised 13lbs.
[None]
[None Given]
Answer
Selected
Answer:
Two-sample T for C1 vs C2
N Mean StDev
SE Mean
C1 7 12.514 0.241 0.091
C2 5 12.5200 0.0837 0.037
Difference = mu (C1) - mu (C2)
Estimate for difference: -0.0057
95% CI for difference: (-0.2386, 0.2272)
T-Test of difference = 0 (vs not =): T-Value = -0.06 P-Value = 0.955 DF = 7
The Standard deviation of the tablets prior to process
improvement is 0.241 making the variance 0.058081. The
tablets after the process have standard deviation of
0.0837 making the variance 0.00700569.
0.058081 > 0.00700569
the hypothesis that one
variance is greater than the other is true. The
improved process lowered the variance.
Correct
Answer:
[None]
Response
Feedback:
[None Given]
Thursday, May 8, 2014 5:55:42 PM EDT


LSL = 19800
USL = 20800
N = 5000
𝑍𝐿𝑆𝐿 =
𝑍𝐿𝑆𝐿 =
𝐿𝑆𝐿 − µ
𝜎
19800 − 20000
80
𝒁𝑳𝑺𝑳 = −𝟐. 𝟓 = 𝑷𝑳𝑺𝑳 =0.00621
𝑍𝑈𝑆𝐿 =
𝑍𝑈𝑆𝐿 =
𝑈𝑆𝐿 − µ
𝜎
20800 − 20000
80
𝑍𝑈𝑆𝐿 = 10 = 𝑃𝑈𝑆𝐿 = 1
𝐏𝐔𝐒𝐋 = 𝟏 − 𝐏𝐔𝐒𝐋 = 𝟎
Pfailure = PUSL + PLSL
Pfailure =0+0.00621
𝐏𝐟𝐚𝐢𝐥𝐮𝐫𝐞 = 0.00621
n𝐟𝐚𝐢𝐥 = 𝐧(𝐏𝐟𝐚𝐢𝐥𝐮𝐫𝐞 )
n𝐟𝐚𝐢𝐥 = 𝟓𝟎𝟎𝟎(𝟎. 𝟎𝟎𝟔𝟐𝟏)
𝒏𝒇𝒂𝒊𝒍 =31.05
The following is the minitab result
Cumulative Distribution Function
Normal with mean = 20000 and standard deviation = 80
x
P( X <= x )
19800
0.00621
20800
1.00000
N = 100
P = 0.10
X=8
𝑛
p(x) = ( ) 𝑝 𝑥 (1 − 𝑝)𝑛−𝑥
𝑥
n!
p(x) =
𝑝 𝑥 (1 − 𝑝)𝑛−𝑥
𝑥! (𝑛 − 𝑥)!
100!
p(x) =
0.108 (1 − 0.10)100−8
8! (100 − 8)!
p(x) = 0.1148230
11.5%
The following is the minitab result
Probability Density Function
Binomial with n = 100 and p = 0.1
x
P( X = x )
8
0.114823