Econometrics and Business Statistics - Solved Assignments

This text provides solved assignments for Econometrics and Business Statistics course covering topics like probability, hypothesis testing, regression analysis and more. It includes descriptive statistics, F-test, two-sample t-test, scatter plot, regression model and significance level analysis.

ECONOMETRICS AND BUSINESS STATISTICS
STUDENT NAME/ID
[Pick the date]

Question 1
(a) Average numeracy score achieved by year 5 classes = 493
Standard deviation = 60
Randomly selected sample size = 30
Probability that a randomly selected sample would have average numeracy score less than 500
=?
Standard error (S.E.) = Standard deviation/ sqrt (Randomly selected sample size)
S . E .= 60
√ 30 =10.95
Assumption: It has been assumed that the mean numeracy scores has followed normal
distribution.
Standard Normal Table
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Probability that a randomly selected sample would have average numeracy score less than 500 is
0.7389.
(b) Assuming that 50% of student scored higher than 493 in numeracy test and hence, P(o) =
0.50
Randomly selected sample size N = 200
Total number of students scored higher than 493 = 115
Relevant proportion p= 115
200 =0.575
The z score= p− p(o)
√ p ( o )∗1− p ( o )
N
Hence, requisite probability P(z>2.1213) = 0.0169 (Standard Normal Table)
(c) Average numeracy score achieved by year 5 classes (2016) = 493
Randomly selected sample size = 59
Standard deviation (2015) = 36.76
Average numeracy score (2015) = 506.63
 Null and alternative hypotheses
Null hypothesis H0 :μ2015=μ2016
Alternative hypothesis H1 :μ2015 ≠ μ2016
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 The t stat
t stat= 493−506.63
36.76
√ 59
=−2.848
 Degree of freedom
(Dof) =Randomly selected sample size-1 =58
 The p value
It is apparent from the sign of alternative hypothesis that it is a two tailed hypothesis test and
hence, the two tailed p value would be taken into account.
The p value = 0.0060
 Level of significance
Alpha = 5%
 Result
It is apparent that p value is lesser than level of significance and therefore, sufficient statistically
evidence is available for the rejection of null hypothesis and for the acceptance of alternative
hypothesis. Hence, the claim is correct that average numeracy scores in 2015 in three major
capital cities differs from the average numeracy scores in 2016.
(d) 95% confidence interval
Randomly selected sample size = 59
Standard deviation (2015) = 36.76
Average numeracy score (2015) = 506.6
The t critical value (95% confidence interval and 59 dof) = 2.0017
Lower limit = Average – (t value * Standard deviation/ sqrt(sample size)) = 506.6 – (2.0017*
36.76/sqrt(59)) = 497.05
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Upper limit = Average – (t value * Standard deviation/ sqrt(sample size)) = 506.6 +(2.0017*
36.76/sqrt(59)) = 516.20
95% confidence interval = [497.05 516.20]
There is 95% confidence that mean numeracy score for grade 5 class of three major capital citiies
in the year 2015 will lie in the range of 497.05 and 516.20.
(e) Assuming that 50% of year 5 classes Melbourne student achieved higher than or same as 500
in numeracy test and hence, P(o) = 0.50
Randomly selected sample size N = 20
Total number of students scored higher than or same as 500 = 15
Relevant proportion p= 15
20 =0.75
 Null and alternative hypotheses
Null hypothesis H0 :μ year 5 ,Melbourne <500
Alternative hypothesis H1 :μ year5 , Melbourne ≥ 500
 The t stat
t score= p− p(o)
√ p ( o )∗1−p ( o )
N
= 0.75−0.5
√ 0.5 ( 1−0.5 )
20
=2.24
 Degree of freedom
(Dof) =Randomly selected sample size-1 =19
The p value
It is apparent from the sign of alternative hypothesis that it is a one tailed hypothesis test and
hence, the two tailed p value would be taken into account.
The p value = 0.0186
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(f) Level of significance
Alpha = 5%
(g) Result
It is apparent that p value is lesser than level of significance and therefore, sufficient statistically
evidence is available for the rejection of null hypothesis and for the acceptance of alternative
hypothesis. Hence, the claim is correct of year 5 classes Melbourne student achieved higher than
or same as 500 in numeracy test.
(f) 95% confidence interval
95% confidence interval = [0.5602 0.9398]
There is 95% confidence that mean numeracy score of year 5 classes Melbourne students
achieved higher than or same as 500 in numeracy test will lie in the range of 0.5602 and 0.9398.
(h) It is apparent that proportion of year 5 classes Melbourne students at 65% fall within the
range of 95% confidence interval and hence, null hypothesis will not be rejected and hence, it
cannot reject the claim that 65% of year 5 classes Melbourne students achieved higher than
or same as 500 in numeracy test.
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Question 2
(a) Descriptive stat (year of 3 classes students from Sydney and Melbourne)
(b) Hypothesis testing : Null and alternative hypothesis
F test – two sample for variance
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Two tailed p value = 2*one tailed p value = 0.85
Level of significance (Alpha) = 5%
It is apparent that p value is greater than level of significance and therefore, sufficient
statistically evidence is not available for the rejection of null hypothesis. Hence, there is no
significant differences is observed in the variance of average writing scores of the year 3 classes
children from Sydney school and Melbourne school in the year 2015.
(c) Hypothesis testing
Null and alternative hypothesis
Population standard deviation is not given which indicates that z value cannot be used and hence,
relevant test would be two sample t test.
(d) T
he
p
value (two tailed) comes out to be 0.9209. It can be seen that p value is greater than alpha
(5%) and therefore, sufficient statistically evidence is not available to reject the null
hypothesis. Thus, no statistically significant different is present in average writing scores of
the year 3 class children from Sydney and Melbourne school in the year 2015.
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(e) Histogram
It can be seen from the above that histogram does not show normal distribution because there is a
long leftward tail which is indicative of presence of negative skew. The histogram also indicates
the variation in the data points.
(f) Hypothesis testing
Population standard deviation is not given which indicates that z value cannot be used and hence,
relevant test would be two sample t test.
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One tailed p value = 0.0173
Level of significance (Alpha) = 5%
It is apparent that p value is lesser than level of significance and therefore, sufficient statistically
evidence is available for the rejection of null hypothesis and acceptance of alternative
hypothesis. Hence, the mean reading scores is higher than mean writing scores of Melbourne
school children of year 3
(g) The 95% confidence interval for difference of ready and writing scores of Melbourne school
children of year 3 is shown below.
95% confidence interval = [5.547 33.553]
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There is 95% confidence that difference of ready and writing scores of Melbourne school
children of year 3 will lie in the range of 5.547 and 33.553.
Question 3
(a) Scatter plot
X: Independent variable= Numeracy scores
Y: Dependent variable= Grammar scores
The association between the variables can be termed as linear because numeracy and grammar
scores are fall on a straight line.
(b) Regression model
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Least square regression equation
Grammar scores = (-4.66) + (1.04*Numeracy scores) + (14.264)
(c) Requisite sample regression model
(d) The slope coefficient coms out to be 1.04 which indicates that for a unit variation in the
numeracy scores would result corresponding change in the grammar scores by 1.04 times.
(e) Significance level = 5%
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Null Hypothesis: slope (βnumeracy) is significant.
Alternative Hypothesis: slope (βnumeracy) is insignificant.
The t stat from the regression for slope coefficient = 7.82
Corresponding p value = 0.00
Level of significance (Alpha) = 5%
It is apparent that p value is lesser than level of significance and therefore, sufficient statistically
evidence is available for the rejection of null hypothesis and acceptance of alternative
hypothesis. Hence, it can be said that slope coefficient is significant which implies that numeracy
scores is a statistically significant predictor of dependent variable grammar score.
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