Statistics Assignment: Hypothesis Testing, Regression, and Analysis

This statistics assignment solution provides detailed answers to questions covering various statistical concepts. It begins with an overview of probability, including classical, relative frequency, and subjective approaches. The solution then delves into hypothesis testing, using paired t-tests and ANOVA to analyze data and draw conclusions. Regression analysis is also explored, with the derivation of regression equations, estimation of values, and calculation of correlation coefficients. The assignment further includes analysis of variance (ANOVA) and the interpretation of statistical outputs, such as F-tests and p-values, to determine the significance of different variables. Overall, the document offers a comprehensive guide to solving statistical problems and understanding key statistical principles.

STATISTICS

Table of Contents
Question 1........................................................................................................................................3
a. Types of probability assigning methods..................................................................................3
b.i.................................................................................................................................................3
b.ii................................................................................................................................................3
Question 2........................................................................................................................................3
a....................................................................................................................................................3
b...................................................................................................................................................3
Question 3........................................................................................................................................3
a....................................................................................................................................................3
b...................................................................................................................................................3
c....................................................................................................................................................4
d...................................................................................................................................................4
e....................................................................................................................................................4
Question 4........................................................................................................................................4
a....................................................................................................................................................4
b...................................................................................................................................................4
c....................................................................................................................................................5
d...................................................................................................................................................7
Question 5........................................................................................................................................7
i. Derivation of regression equation.............................................................................................7
ii...............................................................................................................................................8
iii..................................................................................................................................................9
Question 6......................................................................................................................................10
a..................................................................................................................................................10
b.................................................................................................................................................11
c..................................................................................................................................................11

Question 1
a. Types of probability assigning methods
The types of probability assigning methods have been described in the form of approaches:
1. Classical approach: Discusses about equally likely events
2. Relative frequency: Assigning of probabilities on the behalf of experimentation or historical
data.
3. Subjective approach: Assignment of probabilities based on subjective judgment.
b.i.
Percentage of the members felt that both job security and salary increment were important
= 0.65 * 0.60 = 0.39
Hence, 39% members feel both job security and salary increment important.
b.ii.
Out of 100%; 39% feel both job security and salary increment important. Hence, remaining feel
at least on these two issues was important. So, percentage of such members is:
= 100 – 39 = 61%
Question 2
a.
The value 12 reflects meals per month, the value of standard deviation = 12 – 10 = 2. And 95%
of values are within 2 standard deviation of the mean (68-95-99.7 rule). Thus, out of 100; 95
students consume more than 12 instant meals per month.
b.
Consumption of instant meals per student = 275 / 25 = 11. And 11 is the indication of 1 standard
deviation (11 – 10 = 1). Where, 68% of values are within 1 standard deviation (68-95-99.7 rule)
and 100% are more than 1 standard deviation of the mean. So, there is 100% chance that a
random sample of 25 students will consume more than 275 instant meals.

Question 3
a.
H0: There is no significance difference between the mean of pre and post installation of the
safety equipment.
H1: There is significance difference between the mean of lost pre and post installation of the
safety equipment.
b.
The suitable test statistics here will be paired t-test. The reason is, paired t-test measures the
mean of two identical samples to find whether there is any difference between the mean of both
samples. If there is any difference in the mean indicates that there is variation in the result due to
installation of safety equipment.
c.
Sample mean (mean difference) = -1.2 (given in the question)
Formula of Paired t-test:
t =
d
√ S 2
n
Where d bar is the mean difference, s² is the sample variance, n is the sample size and t is a
Student t quartile with n-1 degrees of freedom.
Standard deviation = 5 (given in the question)
Sample variance s2 = σ 2 = 52 = 25
N = 50 (given in the question)
t = 1.2
√25 /50
t = 1.69
Degree of freedom = n – 1 = 50 -1 = 49
Critical value of t (0.90; 49) = 1.299 (From t table)
P value = 0.10.

d.
The decision criteria is: if the value of t is greater than 1.299 than null hypothesis will be
rejected, and if the value of t is lower than 1.299 (t critical value) then null hypothesis will be
accepted and it will indicate no difference between mean of pre post data collected.
e.
The value of t is greater than t critical value (1.69 > t > 1.299).
Thus, null hypothesis will be rejected with acceptance of alternative hypothesis.
It can be concluded that there is significance difference between mean of pre and post
installation. And it also indicates the effectiveness of equipment, as it can reduce the number of
person hours lost.
Question 4
a.
H0: There is no significance difference in the perception among three groups
H1: There is significance difference in the perception among three groups
b.
The decision criteria will be based on p value.
If the p value is greater than 0.05 which is a significance level; then it will be failed to reject null
hypothesis. On the other hand, if p value will be lower than 0.05 then null hypotheses will
successfully rejected.
c.
Test statistic
A B C
6 5 6
5 5 7
4 4 6
5 4 5
6 5 6
4 4 6
5 5 6

4 6 6
6 5 4
5 6 5
∑X = ∑A + ∑B + ∑C
∑A = 6 + 5 + 4 + 5 + 6 + 4 + 5 + 4 + 6 + 5 = 50
∑B = 5 + 5 + 4 + 4 + 5 + 4 + 5 + 6 + 5 + 6 = 49
∑C = 6 + 7 + 6 + 5 + 6 + 6 + 6 + 6 + 4 + 5 = 57
Ethical values
A B C Total
N 10 10 10 30
∑X 50 49 57 156
Mean
(∑X/N)
5 4.9 5.7 5.2
SSB = 10 (5 – 5.2)2 + 10 (4.9 – 5.2)2 + 10 (5.7 – 5.2)2
= 0.4 + 0.9 + 2.5 = 3.8
SSE:
A A - 5 (A-5)2
6 1 1
5 0 0
4 -1 1
5 0 0
6 1 1
4 -1 1
5 0 0
4 -1 1
6 1 1
5 0 0
6

B B - 4.9
(B -
4.9)2
5 0.1 0.01
5 0.1 0.01
4 -0.9 0.81
4 -0.9 0.81
5 0.1 0.01
4 -0.9 0.81
5 0.1 0.01
6 1.1 1.21
5 0.1 0.01
6 1.1 1.21
4.9
C C - 5.7 (C-5.7)2
6 0.3 0.09
7 1.3 1.69
6 0.3 0.09
5 -0.7 0.49
6 0.3 0.09
6 0.3 0.09
6 0.3 0.09
6 0.3 0.09
4 -1.7 2.89
5 -0.7 0.49
6.1
SSE = 6 + 4.9 + 6.1 = 17
Result Details
Source SS df MS
Between-ethical values 3.8 2 3.8/2 = 1.9 (1.9/0.6296) =

Within- ethical values 17 27 17/27 = 0.6296 F = 3.01765
Total 20.8 29
The f-ratio value is 3.01765. The p-value is 0.065646.
d.
The value of p is greater than 0.05. Thus, the result is not significant at p < .05. Hence, there is
no significant difference in the mean price of gasoline for three bands.
Question 5
i. Derivation of regression equation
X Values Y Values
3.5 100
3 120
2.8 150
3.6 130
2.75 170
3.4 135
3.12 130
2.86 185
3.02 127
2.6 190
3.3 96
∑ = 33.95 1533
M: 3.0864 M: 139.3636
X - Mx Y - My (X - Mx)2 (X - Mx)(Y - My)
0.4136 -39.3636 0.1711 -16.2822
-0.0864 -19.3636 0.0075 1.6723
-0.2864 10.6364 0.082 -3.0459
0.5136 -9.3636 0.2638 -4.8095
-0.3364 30.6364 0.1131 -10.305
0.3136 -4.3636 0.0984 -1.3686
0.0336 -9.3636 0.0011 -0.315
-0.2264 45.6364 0.0512 -10.3304

-0.0664 -12.3636 0.0044 0.8205
-0.4864 50.6364 0.2365 -24.6277
0.2136 -43.3636 0.0456 -9.264
SS: 1.0749 SP: -77.8555
Sum of X = 33.95
Sum of Y = 1533
Mean X = 3.0864
Mean Y = 139.3636
Sum of squares (SSX) = 1.0749
Sum of products (SP) = -77.8555
Regression Equation = ŷ = bX + a
b = SP/SSX = -77.86/1.07 = -72.43348
a = MY - bMX = 139.36 - (-72.43*3.09) = 362.91969
ŷ = -72.43348X + 362.91969
ii.
Estimation of y, when x is 2.5%
ŷ = -72.43348X + 362.91969
Putting value of x in given equation;
ŷ = -72.43348 (2.5) + 362.91969
y = 181.84
Hence, at mortgage interest of 2.5%, the estimation number of housing starts will be 189.

iii.
X Values Y Values
3.5 100
3 120
2.8 150
3.6 130
2.75 170
3.4 135
3.12 130
2.86 185
3.02 127
2.6 190
3.3 96
X - Mx Y - My (X - Mx)2 (Y - My)2 (X - Mx)(Y - My)
0.414 -39.364 0.171 1549.496 -16.282
-0.086 -19.364 0.007 374.95 1.672
-0.286 10.636 0.082 113.132 -3.046
0.514 -9.364 0.264 87.678 -4.81
-0.336 30.636 0.113 938.587 -10.305
0.314 -4.364 0.098 19.041 -1.369
0.034 -9.364 0.001 87.678 -0.315
-0.226 45.636 0.051 2082.678 -10.33
-0.066 -12.364 0.004 152.86 0.82
-0.486 50.636 0.237 2564.041 -24.628
0.214 -43.364 0.046 1880.405 -9.264
Mx: 3.086
My:
139.364
Sum:
1.075
Sum:
9850.545 Sum: -77.855
X Values
∑ = 33.95
Mean = 3.086
∑(X - Mx)2 = SSx = 1.075
Y Values
∑ = 1533

Mean = 139.364
∑(Y - My)2 = SSy = 9850.545
X and Y Combined
N = 11
∑(X - Mx)(Y - My) = -77.855
R Calculation
r = ∑((X - My)(Y - Mx)) / √((SSx)(SSy))
r = -77.855 / √((1.075)(9850.545)) = -0.7566
The value of r is – 0.7566, this indicates that there is negative correlation between interest rate
and number of housing starts. Also the value of r is near to – 1, hence there is strong relationship
between both variables.
Question 6
a.
A
Multiple R = 0.442 (given in the question)
R2 = Multiple R2 = 0.4422 = 0.1953
A = 0.195
B
Df = (Total variables (n) – 1)
Df = (4 – 1) = (4 – 1) = 3
C
C = Residual Df + B
= 21 + 3 = 24
D

D = Residual Df × Residual MS
= 21 × 6.6933 = 140.55
D = 140.55
E
E = Regression SS / B
E = 34.1036 / 3 = 11.368
F
F = E / Residual MS
F = 11.368 / 6.6933 = 1.698
G
G = Television Standard Error × Television t Stat
G = 1.96 × 0.37 = 0.7252
H
H = Direct Coefficients / Direct Standard Error
H = 0.57 / 1.72 = 0.3313
b.
The value of F is lower than its significance value (1.698 < 0.1979)
Thus, the independent variable is significant at 5% level.
c.
The two variables; Direct (0.74) and Television (0.71) have P-value more than 0.05; thus model
is significant at 5% level. Only Newspaper carries P value below 0.05 (0.04), thus it is non-
significant variable.