Statistics Assignment: Confidence Intervals and Regression

This document presents a comprehensive solution to a statistics homework assignment, addressing key statistical concepts and methodologies. The solution encompasses calculations and interpretations for confidence intervals, determining sample sizes, and performing hypothesis tests using Z-scores and p-values. It also includes detailed analysis of regression models, deriving regression equations, interpreting slopes and coefficients of determination, and testing the significance of relationships between variables. The assignment covers topics such as the relationship between salaries and population, production line efficiency, and wealth and age correlation. The document provides step-by-step calculations, decision rules, and conclusions for each question, demonstrating a strong understanding of statistical principles and their application in business contexts. The content is designed to aid students in understanding and applying statistical techniques to real-world problems.

Strategics for business
Decision

Table of Contents
Week 7: Question 2..........................................................................................................................1
a. Provide 95% confidence interval for all patients....................................................................1
b. Sample size required to estimate proportion of all hospitals..................................................1
Week 8: Question 10........................................................................................................................2
Hypothesis...................................................................................................................................2
Test statistic formula...................................................................................................................2
Level of significance...................................................................................................................2
Decision rule...............................................................................................................................2
Calculation..................................................................................................................................2
Conclusion...................................................................................................................................3
Week 9: Question 3..........................................................................................................................3
a. Null and alternative hypothesis...............................................................................................3
b. Decision rule ..........................................................................................................................3
c. Calculate the test statistic........................................................................................................3
d. Decision...................................................................................................................................5
Week 10: Question 2........................................................................................................................5
a. Estimated regression line and interpret the slope....................................................................5
b. Estimated total personal wealth when a person is 50 year old................................................5
c. What is value of coefficient of determination.........................................................................6
d. testing the value at 10% level of significance.........................................................................7
Week 11: Question 3........................................................................................................................8
a. Estimated regression equation.................................................................................................8
b. Compute the coefficient of determination...............................................................................9
c. Test to determine y is significantly related to independent variable.......................................9
d. Test whether x and y are significantly related......................................................................10

Week 7: Question 2
a. Provide 95% confidence interval for all patients
The proportion is as mention below:
Step 1:
CI = p +- Z(a/2) √p (1 – p) / n
in which P = sample proportion and n= sample size
So, as per the question n= 400 and the level of significance is 0.05 which is fixed.
Therefore, sample proportion = x / n
So, sample proportion = 800 / 400
= 0.2
Step 2:
In accordance with the table of standard normal distribution, there is a need to find the critical
value and the level of significance is 0.05 in which z= 1.96
so, = CI = p +- Z(a/2) √p (1 – p) / n
=0.2 +,- 1.96 √0.2 (1 – 0.2) / 400
= 0.2 +- 1.96 (0.02)
= 0.2 +- 0.0392
= (0.1608 and 0.2392)
b. Sample size required to estimate proportion of all hospitals
Confidence level 95%
Margin of error 4%
Population Proportion 50%
Sample size 601
So, it is interpreted that when 0.04% of margin of error is given, then the sample size estimated
for proportion of all hospitals referrals to the health care is 601.
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Week 8: Question 10
Hypothesis
H>0 Null Hypothesis: To determine there is no significant relationship between salaries and
population
H>1 Alternative hypothesis: To determine there is a significant relationship between salaries and
population
Test statistic formula
Z score = x- μ/ Standard deviation
Where x is raw score and is population mean.
Level of significance
The Z score value is 0.2 and on the other hand p value is 0.079 which is greater than 0.05 and
when the value if higher than level of significance which means alternative hypothesis is
accepted. So, it is reflected that there is a significant relationship between salaries and
population.
Decision rule
As per the decision rule, it is stated that if the test statistic follows a normal distribution
then critical value from the standard normal distribution which is known as Z statistic. So the
rule state that when the Z score value of greater than 0.05 then the alternative hypothesis is
accepted and on the other other side when the value is less than 0.05, if means the null
hypothesis is accepted while alternative hypothesis is rejected.
Calculation
Step 1: x- μ/ Standard deviation
Step 2: 50000-48400 / 8000
Step 3: 0.2
So, p-value from the Z table shows that
P (x<50000) = 0.57926
P(x>50000) = 1- P (x<50000) = 0.42075
P (48400 < x < 50000) = P (x< 50000) – 0.5 = 0.07926
2

Conclusion
So, from the above, it has been concluded that there is a relationship between the salaries
and population because the value is greater than 0.05. So, it can be stated that there is an
increase in the starting salaries when the employees are hire as compared to after some time.
Week 9: Question 3
a. Null and alternative hypothesis
H0: Null hypothesis: To analyze there is no relationship between average value of production of
three lines and hourly quantity
H1: Alternative hypothesis: To analyze there is a relationship between average value of
production of three lines and hourly quantity
b. Decision rule
As per the decision rule, it is analyzed that there are two rules which proves the testing of
an hypothesis such that,
 When the value of p which is greater than 0.05, in that case null hypothesis is rejected
while alternative hypothesis is accepted. (P > 0.05 = alternative hypothesis)
 On the other side, the value of p is less than 0.05, so null hypothesis is accepted and on
the other side, alternative hypothesis is rejected (P < 0.05= null hypothesis).
c. Calculate the test statistic
Group Count Sum Average Variance
Process 1 4 120 30 4.66
Process 2 4 136 34 11.33
Process 3 4 128 32 13.33
Anova
Source of
variation
SS df MS F P value
Between
Groups
32
(SST)
2 16 1.63 0.24
3

Within Groups 88
(SSE)
9 9.77
Total 120 11
Step 1: to determine the value of df
df= k-1
where k = total number of group
so, here total number of groups are 3 which process 1, process 2 and process 3
df= 3-1
df = 2
Step 2: to determine df value within a groups,
so, df for second source = N-k
here N is total number of process in anova table which is 12
So, df = 12 – 3
= 9
Therefore total value is 9+2 = 11
Step 3: To determine the value of MS, the below mention formula is used i.e.
For first source, MS = SST / (k – 1)
MS = 32 / (3-1)
MS= 32 / 2
MST = 16 (1st source)
Next for second source = MS = SSE / (N – k)
MS = 88 / (12 – 3)
MS = 88 / 9
MSE = 9.77
Step 3: To determine the value of F, the formula is used which is as mention below:
F = MST / MSE
F = 16 / 9.77
F = 1.63
4

Step 4: to determine the value of p- significant value
= the value if f- distribution of 1.63 is 0.24
d. Decision
In accordance with the above table, it is interpreted that the value of significant factor is
greater than 0.05 i.e. (P > 0.05) that means alternative hypothesis is accepted and on the other
side, rejecting the null hypothesis. So, the decision reflected that there is a significant
relationship between the mean value of three lines of production process with the hourly
quantity of production within a XYZ company. So when the working hours within a firm is
increases it means that the production of all the three lines are also increases. Therefore, this
shows a positive relationship between both the variables in the quoted organization.
Week 10: Question 2
a. Estimated regression line and interpret the slope
The regression line equation is ŷ= a + bx, here y is the dependent variable and x is an
independent variable.
Through the above graph, it is interpreted that there is a relationship between both
variables I.e independent and dependent variables. So it is stated that with when the age of
person is increases, there is an increase in chances of wealth as well. This reflect the positive
interrelationship between the two variables.
5
30 40 50 60 70 80 90
0
100
200
300
400
500
Age Line Fit Plot
Column B
Column V
Age
Total wealth

b. Estimated total personal wealth when a person is 50 year old
X 50
Regression equation 5.3265*x+45.2159
266.325
Total personal wealth 311.5409
c. What is value of coefficient of determination
SUMMARY
OUTPUT
Regression Statistics
Multiple R 0.9547040663
R Square 0.9114598542
Adjusted R Square 0.8967031633
Standard Error 28.9895420825
Observations 8
ANOVA
df SS MS F
Significance
F
Regression 1
51907.6386
990981
51907.6386
990981
61.7658699
187
0.00022451
52
Residual 6
5042.36130
09019
840.393550
1503
Total 7 56950
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Coefficients
Standard
Error t Stat P-value Lower 95% Upper 95%
Lower
95.0%
Upper
95.0%
Intercept
45.2159059
852
39.8049868
594
1.13593570
94
0.29931074
67
-
52.1833880
957
142.615200
0662
-
52.183388
0957
142.61
52000
662
Age
5.32659196
5
0.67775877
39
7.85912653
66
0.00022451
52
3.66817598
91
6.98500794
1
3.6681759
891
6.9850
07941
The coefficient of determination is the value of R square and the value is 0.911 which is
approx 0.1. so it can be stated that there is a positive linear association and the dependent value
is 10% fluctuate in positive or negative side of independent value.
d. testing the value at 10% level of significance
At 90% Confidence level
Step 1 : Hypothesis
H0: Null Hypothesis: To determine there is no relationship between wealth and age
H1 : Alternative Hypothesis: To determine there is a relationship between wealth and age
Step 2 : Standardized the test statistics
SUMMARY OUTPUT
Regression Statistics
Multiple R 0.9547040663
R Square 0.9114598542
Adjusted R Square 0.8967031633
Standard Error 28.9895420825
Observations 8
Step 3: Level of significance
the level of significance is known as alpha which is clearly shows that probability of
rejecting the null hypothesis or alternative hypothesis and in the case the level of confidence is
90% because 10% level o significance is chosen. So as per the decision rule, decision is taken in
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order to determine there is actual difference or not. So, in accordance with the anova table it is
interpreted that the value of p is 0.00 which is greater than 0.01 and that is why, alternative
hypothesis is accepted.
Step 4: Decision rule
The hypothesis testing rule shows that when the value of level of significance is greater
than 0.01 than alternative hypothesis is accepted and it is so because the level to check
confidence is 90% and on the other side, when the value is less than 0.01 then null hypothesis is
accepted while alternative hypothesis is rejected.
Step 5: Calculation
ANOVA
df SS MS F
Significanc
e F
Regression 1
51907.6
3869909
81
51907.638699
0981
61.76586
99187
0.00022451
52
Residual 6
5042.36
1300901
9
840.39355015
03
Total 7 56950
Coefficients
Standard
Error t Stat P-value Lower 95% Upper 95%
Lower
90.0%
Upper
90.0%
Intercept
45.2159059
852
39.8049868
594
1.13593570
94
0.29931074
67
-
52.1833880
957
142.615200
0662
-
32.132359
5461
122.56
41715
166
Age
5.32659196
5
0.67775877
39
7.85912653
66
0.00022451
52
3.66817598
91
6.98500794
1
4.0095844
807
6.6435
99449
3
8

Step 6: Conclusion
As per the above table, it is concluded that the level of significance or p value is greater
than 0.01 i.e. 0.00 and that is why, alternative hypothesis is accepted and as a result, it is stated
that there is a close relationship between the wealth and age of person. Like, when the age of a
person increases, there is a definitely increase in the wealth.
Week 11: Question 3
a. Estimated regression equation
The estimated regression line is y= b0 + b1x1 + b2x2 + b3x3
In this, y is dependent variable while x1, x2 and x3 are independent variables
So, with the help of data provided, the equation is
b0 = 0.0136, b1 = 0.7992, b2 = 0.2280 and b3 = -0.5796
therefore, the regression equation is
y = 0.0136 + 0.7992x1 + 0.2280x2 - 0.5796x3.
b. Compute the coefficient of determination
The coefficient of determination is also known as R square in which different rules are
applicable. Also, as per the given situation, the coefficient of determination is
Regression sum of squares is 45.936 while on the other side, error sum of squares is
2.6218
Therefore, the coefficient of determination is as mention below:
R square = SSR / SSR +SSE
= 45.9634 / 45.9634 + 2.6218
= 45.9634 / 48.5852
= 0.9460
So, the value is positive and high, so it is reflected that the relationship between the variables is
strong.
c. Test to determine y is significantly related to independent variable
For testing variable x1
H0: Null hypothesis: There is no significant relationship between family spending and income
9

H1: Alternative hypothesis: There is a significant relationship between family spending and
income.
To determine the value of p the formula is used
= Coefficient / standard error
= 0.7992 / 0.074
= 10.8
The p value for the x1 is 1 which is greater than 0.05 and this shows that the alternative
hypothesis is accepted and null hypothesis is rejected.
For testing variable x2
H0: Null hypothesis: There is no significant relationship between family spending and family
size
H1: Alternative hypothesis: There is a significant relationship between family spending and
family size.
= Coefficient / standard error
= 0.2280 / 0.190
= 1.2
The p value for this variable is 0.88 which is greater than 0.05 and this is clearly shows that the
alternative hypothesis is accepted and that is why, it is stated that y independent variable i.e.
family spending is directly related to family size.
d. Test whether x and y are significantly related
For testing variable x3
H0: Null hypothesis: There is no significant relationship between family spending and addition
of savings
H1: Alternative hypothesis: There is a significant relationship between family spending and
addition of savings.
To determine the p value, formula is used
= Coefficient / standard error
= -0.5796 / 0.920
= -0.63
10

Here, the p value for this is 0.2651 which clearly reflected that it is greater than 0.05 and that is
why, it is stated that there is x3 and y are directly related or it has a significant relationship.
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