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Statistics Exam

STATISTICS ASSIGNMENT Page 1 of 13
Statistics Assignment
Student Name
Institutional Affiliation
STATISTICS ASSIGNMENT Page 2 of 13
Instructions:Be sure to record the final answer next to “Answer” and show all your calculation
steps or assumptions under “Work.” You can type in your calculations, use the equation tool.
1. You choose an alpha level of .01 and then analyze your data.
A. What is the probability that you will make a Type I error given that the null hypothesis
is true?
B. What is the probability that you will make a Type I error given that the null hypothesis
is false?
Answer: a) 0.01, b) 0.99
Work: (a) the probability of accepting the null hypothesis when it is true is given by p=α
=0.99
(b) The probability of rejecting the null hypothesis when it is true is given by p=1-α=
0.99
2. a.True/false: You accept the alternative hypothesis when you reject the null hypothesis.
b. True/false: You do not accept the null hypothesis when you fail to reject it.
Answer: a) True b) False
Work: a) Rejecting the null hypothesis implies accepting the alternative hypothesis. That is,
there is not sufficient evidence to support the null hypothesis.
b) failing to reject null hypothesis implies accepting it.
STATISTICS ASSIGNMENT Page 3 of 13
3.
Answer:a)
Subject 1: (-1)(4) + (0)(6) + (1)(7) = 3
Subject 2: (-1)(3) + (0)(7) + (1)(8) = 5
Subject 3: (-1)(2) + (0)(8) + (1)(5) = 3
Subject 4: (-1)(1) + (0)(4) + (1)(7) = 6
Subject 5: (-1)(4) + (0)(6) + (1)(9) = 5
Subject 6: (-1)(2) + (0)(4) + (1)(2) = 0
b) t calculated = 4.158, t critical=0.0088 hence we can conclude that the subjects are
getting better each trial.
Work: a)
A
b
c
L
STATISTICS ASSIGNMENT Page 4 of 13
4
6
7
-1*4 +0*6+7*1=3
3
7
8
-1*3 +0*7 *1*8=5
2
8
5
-1*2+0*8+1*5 =3
1
4
7
-1*1 +0*4 +1*7 =6
4
6
9
-1*4 * 0*6 + 1*9 =5
2
4
2
-1*2 + 0*4 +1*2 =0
b) Sample mean, x
̅
= (3 + 5 + 3 + 6 +5 +0)/6
=3.667, n=6
Standard deviation, s= 2.160
t-statistic=
̅

= 3.667/ (2.160√6) = 4.158
T critical=t
0.05,5
= 2.02
4. You are conducting a study to see if students do better when they study all at once or in intervals.
One group of 12 participants took a test after studying for one hour continuously. The other group
of 12 participants took a test after studying for three twenty minute sessions. The first group had a
mean score of 75 and a variance of 120. The second group had a mean score of 86 and a variance
of 100.
a. What is the calculated t value? Are the mean test scores of these two groups significantly
different at the .05 level?
STATISTICS ASSIGNMENT Page 5 of 13
b. What would the t value be if there were only 6 participants in each group? Would the scores be
significant at the .05 level?
Answer: a) Mean test scores of the two groups is significantly different at 0.05 significance
level
b) The scores remain significantly different at 0.05 significance level even after reducing the
sample size.
Work: a) t calculated =
µµ
²

²

=





|t calculated|= 8.325
t-value= 8.325
p-value is <0.00001 which is less than 0.05 hence the scores are significantly different
T critical
0.05, 23
= 1.71
The t calculated value is greater than the t critical value hence we conclude that the scores are
significantly different at alpha=0.05
b)
µµ
²

²

=



= |t calculated|= 5.886
t-value= 5.886
p-value = 0.000053
T critical
0.05, 11
= 1.78
Reducing the sample size does not affect the results and as such the mean test scores of the two
groups are significant at alpha =0.05
STATISTICS ASSIGNMENT Page 6 of 13
5.Note: In order to answer the question, you don’t actually need to conduct the hypothesis
test as the question asks.
Answer: a) H
0
: µ= 4.75, H
a
: µ> 4.75
Work


=


= -0.4841
t-value =-0.4841
t
tabulated value
= 2.145
the t calculated value is less than the t tabulated value. Hence we fail to reject the null
hypothesis and conclude that the sample mean hours are greater than 4.75
STATISTICS ASSIGNMENT Page 7 of 13
6.
Answer:b. to conclude that the current mean hours per week is higher than 4.5, when in fact,
it is the same.
Work:
a) H
0
: current mean hours per week is equal to 4.5
H
1
: current mean hours per week is higher than 4.5
Critical region: Reject H
0
whenever t
calculated=


is greater than t critical.
µ=4.5, x
̅
= 4.75, n=15, s=2.0
|t|= 4.5-4.75/ (2.0/√15)
=0.4841
T
0.05,14
= 1.76
T calculated value is less than t critical hence we reject the null hypothesis and conclude that the
current mean hours is higher than 4.5
7. An article in the San Jose Mercury News stated that students in the California state university
system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that
STATISTICS ASSIGNMENT Page 8 of 13
the mean time is longer. You conduct a survey of 49 students and obtain a sample mean of 5.1
with a sample standard deviation of 1.2. Do the data support your claim at the 1% level?
Answer: the data supports my claims that the mean time taken by students to finish their
undergraduate studies is longer than 4.5 years.
Work:
Hypotheses: H
0
: µ
1
=µ
0
H
1:
µ
1
≥µ
0
Critical region: Reject H
0
whenever t
calculated=


is greater than t critical.
µ=4.5, x
̅
= 5.1, n=49, s=1.2
t
calculated
= 4.5-5.1/ (1.2/7)
= -3.5
Absolute t
calculated value
= 3.5
T critical=t
0.1,48
= 2.39
t-value= 2.39
p-value= 0.001015
p-value is less than 0.1 hence we reject the null hypothesis and conclude that the mean time from
my survey is longer.
The t calculated value is greater than the t critical value. We hence reject the null hypothesis and
conclude that the mean time from my survey is longer.
STATISTICS ASSIGNMENT Page 9 of 13
8.
Answer: The mean difference between the girls jumping and relaxed times is zero.
Work:
Hypotheses: H
0
: µ
1
- µ
2
≠ 0
H
1
: µ
1
- µ
2
=0
STATISTICS ASSIGNMENT Page 10 of 13
Critical region: Reject the null hypotheses whenever Z=



²

is greater than αthe
significance level.
x
̅
1
= 259/8=32.375 x
̅
2
= 245/8=30.625
1
= 647.875/7=92.5536
2
=485.875/7= 69.4107
Z calculated= 32.375-30.625/ (√92.5536/8 +69.4107/8)
=0.3889
p-value= 0.69735. the p value is greater than alpha the significance level hence it is not significant
at 0.05.
Assuming α= 0.05, then Z calculated is greater than alpha. We therefore reject the null hypothesis
and conclude that the mean difference between the girls jumping times is zero.
9. A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. Of interest
is whether the liquid diet yields a higher mean weight loss than the powder diet. The powder diet
group had a mean weight loss of 42 pounds with a standard deviation of 12 pounds. The liquid diet
group had a mean weigh loss of 45 pounds with a standard deviation of 14 pounds.
Answer:T critical is greater than t obtained we hence fail to reject the null hypothesis and
conclude that the liquid diet yields a higher mean weight loss than the powder diet.
STATISTICS ASSIGNMENT Page 11 of 13
Work:
Pop 1 Pop 2
n
1
= 49n
2
=36
µ
1
= 42 µ
2
=45
s=12 s=14
here the standard deviations are different thus:
hypotheses: H
0
: the powder diet has equal mean weight loss as the liquid diet
H
1
: liquid diet yields a higher mean weight loss than the powder diet
Critical region: Reject H
0
whenever t
obtained
=
µµ
²

²

is greater than t critical
T obtained= 42-45/ (√144/49 +196/36)
= -1.036
T value= 1.036
T critical= t
0.05, n1+n2-2
=1.99
p-value= 0.303211
p value is greater than alpha= 0.05 hence it is not significant.
Conclusion: T critical is greater than t obtained we hence fail to reject the null hypothesis
and conclude that the liquid diet yields a higher mean weight loss than the powder diet.
STATISTICS ASSIGNMENT Page 12 of 13
10.
Answer:
Donot reject the null hypothesis, H
0
Work:
Hypotheses:
H
0
: µ
1
= µ
0,
new technique for improving players golf scores is not effective
H
1
:µ
1
≥ µ
0 ,
new technique for improving players golf scores is effective
Critical region: reject H
0
whenever t=
µµ


is greater than t critical
µ
1
= 85.25 µ
2
= 83
Assuming equal variances, t obtained is calculated by:
T=
µµ


s
1
=40.25, s
2
=12, s
p
= 40.25+12= 52.25
T
obtained
= (85.25-83)/(52.25×0.7071)
STATISTICS ASSIGNMENT Page 13 of 13
=0.0609
t-value= 0.0609
p-value= 0.477634
p-value is greater than alpha =0.05 hence it is not significant
T critical= t
0.025,6
=2.45
Conclusion:
The t obtained values are less than the t critical value. We thus fail to reject the null
hypothesis and conclude that the new technique for improving player’s golf scores is not
effective.

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