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Question 1
U=$51, sigma= $4.20, x bar (sample mean)= $51.85, n= 100 books, alpha= 0.
1. State the H O
μ ≤ $
2. State the H 1
: μ > $
3. Choose Alpha: α =¿0. 4. Choose n: n= 100 books 5. Choose Test : One tailed z test for mean ( sigma known) 6. Set up Critical value(s): Z Critical
= 1.645 (=NORM.S.INV (0.95) ) since it is one tailed
and we choose the upper tail since it is on the right tail.
7. Collect Data: U=$51, sigma= $4.20, x bar (sample mean)= $51.85, n= 100 books, alpha=
8. Compute test statistic and p value:
(51.85- 51)/ (4.20/ square root of 100)
Z
Stat
P value= 0 since the z value is too high
9. Make Statistical Decision: Since Z Stat
of 2.023 > Z Critical
of 1.645, Reject the null
hypothesis (H O
10. Express Conclusion: There is sufficient evidence to indicate that the average cost of
textbooks at JSU Bookstore is higher than $51.
Question 2
u= $3400, sigma= $80 per semester, sample mean (x bar)= $3260, n= 50 students, alpha= 0.
1. State the H
O
: μ ≥ $
2. State the H 1
: μ < $
3. Choose Alpha: α =0. 4. Choose n : n= 50 students 5. Choose Test: One tailed Z-test for mean (sigma known) 6. Set up Critical value(s) Z
Critical
= -1.645 (=NORM.S.INV(0.05) ) since we choose alpha
only and we want to know if value is lower (left tail).
7. Collect Data: u= $3400, sigma= $80 per semester, sample mean (x bar)= $3260, n= 50
students, alpha= 0.
8. Compute test statistic and p value:
= 3260-3400/ (80/ square root of 50)
Z
Stat
= -12.378, p value = 0 since the z value is too high
9. Make Statistical Decision: Since Z Stat
of -12.374 is < Z Critical
of -1.645, we reject the null
hypothesis
10. Express Conclusion: There is sufficient evidence to indicate that the average cost of on
campus accommodation is lower than $3400.
Question 3
μ = 940 points , n= 169 students, sample mean (x bar)= 990 points, sample standard
deviation (s)= 140 points
1. State the H O
: μ = 940 points
2. State the H 1
= u ≠ 940 points
3. Choose Alpha:
α =¿
4. Choose n: n=169 students, degrees of freedom (n-1)= 168 students 5. Choose Test; Two tailed t test ( sigma unknown) 6. Set up Critical value(s)= +/- t Critical
= +1.9742, - 1.9742 (T.INV (0.025, 168) ) since alpha=
0.05/2 due to a two tailed test equals 0.025 and degrees of freedom (n-1)= 169-1 so 168
students)
7. Collect Data: Mean= 940 points , n= 169 students, sample mean (x bar)= 990 points,
sample standard deviation (s)= 140 points
8. Compute Test Statistic and P value: T Stat
= (990-940)/ (140/ square root of 169)
T
Stat
= 4.643, p value= 0 since the t statistic is too high.
9. Make Statistical Decision: Among the two critical values for t being -1.9742and +1.9742,
T
Stat
has a value of 4.642 which falls in the rejection region and is higher than +1.9742.
Reject the null hypothesis (H O
10. Express Conclusion: There is sufficient evidence to conclude that the examination scores
are at the same level of 940 points as in previous years.
Question 4
A. p= 0.65, p S
= 165/300= 0.55, n= 300 students, alpha= 0.
B.
1. State the H O
: p
2. State the H 1
: p > 0.
3. Choose Alpha:
α =0.
4. Choose n: n= 300 students 5. Choose Test: One tailed z test for proportion 6. Set up Critical value (s): Z Critical
= 1.645. (=NORM.S.INV(0.95) ) since we are using alpha of
0.05 so we need to know the value which is higher on the right tail.
Z
Stat
= (p 1
Z
Stat
= 0.767 – 0.792 / Square root of 0.784 (1-0.784) (1/30 + 1/72)
= -0.025/ square root of (0.784 (1-0.784) (1/30 + 1/72)
Z
Stat
9. Since the Z Stat
of -0.2801 does not fall in the region of interest and it is less than the value of
Z
Critical
-2.58,. Do not reject the Null Hypothesis (H O)
10. There is insufficient evidence to conclude that Dr Smith’s success rate is worse that Dr
Taylor’s success rate.
Question 6
1. Null Hypothesis: H O
= p 1
-p 2
2. Alternative Hypothesis: H 1
= p 1
-p 2
3. Choose Alpha:
α =0.
4. Choose n: n 1
= 57 students, n 2
= 100 students
5. Choose Test: Two tailed test for two population proportions 6. Set up Critical Value(s)= +/- 1.96 (NORM.S.INV (0.025)) because alpha/2 of 0.05/2 is
0.025 as this is a two tailed test
7. Collect Data: Assuming 1 is the class of 2020 receiving MMR vaccine and 2 is class of JSU
Students from class of 2010
P
2020
= 43/57= 0.754 P
2010
N
1
= 57 students N 2
= 100 students
X
1
= 43 students X 2
= 83 students
8. Compute the Test Statistic and p value:
Calculate the z stat
= 0.754- 0.803/ square root of 0.803 (1-0.803) (1/57 + 1/100)
9. Make Statistical Decision: The value of Z Stat
of -0.7424 is between +1.96 and -1.96. Do not
reject the null hypothesis (H O
10. Express Conclusion: There is insufficient evidence to conclude that the MMR Vaccination
rate for the two groups is different.
Question 7
1. State the H O
: σ 1
2
= σ 2
2
2. State the H 1
: σ 1
2
≠ σ 2
2
3. Choose alpha: α =0. 4. Choose n: n 1
=11, n 2
= 9, degrees of freedom (n-1)= df 1
= 10, df 2
5. Choose Test:Two tailed F- Test for the ratio of two population variances 6. Set up Critical Value(s): Using excel formula
= F.INV.RT (Alpha/2, df 1
, df 2
= F.INV.RT (0.025, 10,8)
F
Critical
7. Collect Data : JSU Central City and JSU Sea City
We assume 1 is JSU Central City and 2 is JSU Sea City
JSU Central City
JSU Central City
Sample Standard Deviation S1 42.
Sample Size n1 11
Degrees of Freedom df1 10
JSU Sea City
JSU Sea City
Sample Standard Deviation S2 28.
