Statistics Assignment: Statistical Analysis and Regression Models

This statistics assignment solution covers several key statistical concepts. The first question involves constructing a frequency distribution table and a histogram, analyzing the data's skewness and potential outliers, and recommending an appropriate measure of location (median). The second question explores the relationship between demand and unit prices using hypothesis testing, correlation, and the coefficient of determination. The third question employs ANOVA to compare the means of three populations, while the fourth question presents a multiple regression analysis to predict the number of mobile phones sold based on price and advertising spots, including hypothesis testing for the significance of each independent variable. The solution provides detailed calculations, interpretations, and conclusions for each question, making it a valuable resource for students studying statistics.

STATISTICS
ASSIGNMENT
[Pick the date]
Student Name

Question 1
a. Frequency distribution table
(b) Histogram
It is apparent that the shape of the above graph is asymmetric with a skew towards the right
owing to presence of tail on the right. As a result, it is possible that outliers are present on the
right side. Also, the given distribution would not be considered as normal distribution as skew is
present.
(c) Owing to presence of positive outliers, it would be recommended that median must be used
as a measure of location for the given dataset since the mean might get distorted owing to the
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presence of these higher values. However, this is not the case with median which does not take
into consideration the exact values.
Question 2
(a) Whether demand and unit prices are related or not.
Alpha = 0.05
Null hypothesis H0 : β=0
Alternative hypothesis H1 : β ≠ 0
The t value= β
s β
=−2.137
0.248 =−8.6169
The p value for two tailed hypothesis test, t value and degree of freedom = 0.00
It can be seen that p value is lower than level of significance and hence, null hypothesis would be
rejected and alternative would be accepted. Therefore, demand and unit prices are related.
(b) Coefficient of determination
R2= SSR
SST = 5048.818
8181.479 =0.617
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It implies that 61.7% of variation in demand would be explained by corresponding changes in the
unit price.
(c) Coefficient of correlation
R= √Coefficient of determination= √0.617=± 0.786
It is apparent that sign of slope coefficient is negative and hence, the correlation coefficient is -
0.786. It implies strong inverse relation is present between the variables which means that as the
demand increases the unit price would decrease.
Question 3
Total number of observations n= 24
Number of treatments k= 3
Degree of freedom for between treatments = k-1 =3-1 = 2
Degree of freedom for within treatments = n-k =24-3= 21
Null hypothesis H0 : μ1=μ2=μ3
Alternative hypothesis H1 : μ1 ≠ μ2 ≠ μ3
F value from ANOVA table = 25.891
The p value for the F value comes out to be 0.00.
Alpha = 0.05
It can be seen that p value is lower than level of significance and hence, null hypothesis would be
rejected and alternative would be accepted. Therefore, the means of the three populations are not
same and atleast one of them is different.
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Question 4
Number of mobile phones sold per day =y
Price of mobile phones =x1
Number of advertising spots =x2
(a) Regression equation
y=0.8051+ 0.4977 x 1+0.4733 x 2
Number of mobile phones sold per day = 0.8051 + (0.4977 * Price of mobile phones) + (0.4733
* Number of advertising spots)
(b) Alpha = 0.05
SS due to regression = 40.700
SS due to residual = 1.016
n=7 days
Degree of freedom for regression = k = 2
Degree of residual ¿ n−k−1=7−2−1=4
Now,
Null hypothesis H0 : β1=β2=… … … . βk=0
Alternative hypothesis H1 : At least one βi ≠ 0
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F = 80.1181
The corresponding p value is 0.00. It can be seen that p value is lower than level of significance
and hence, null hypothesis would be rejected and alternative would be accepted. Therefore,
significant linear relation is present between the independent variables and dependent variable.
(c) Alpha = 0.05
 For β1
Null hypothesis H0 : β1=0
Alternative hypothesis H1 : β1 ≠ 0
t= b1
sb 1
=0.4977
0.4617 =1.07797
The p value for two tailed hypothesis test, t value and degree of freedom = 0.2060
It can be seen that p value is higher than level of significance and hence, null hypothesis would
not be rejected. Therefore, price of mobile is not significant.
 For β2
Null hypothesis H0 : β2=0
Alternative hypothesis H1 : β2 ≠ 0
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t= b2
sb 2
=0.4733
0.0387 =12.2299
The p value for two tailed hypothesis test and t value = 0.00
It can be seen that p value is lower than level of significance and hence, null hypothesis would be
rejected and alternative would be accepted. Therefore, advertising spots for mobile are
significant to number of phones sold.
(d) The slope coefficient of X2 is 0.4733 which implies as there is an increase in the number
of advertisement spots by one, the corresponding increase in the daily sale of mobile
phones would be 0.4733 units.
(e) Price of mobile x1 = $20,000
Number of advertising spots x2 = 10
y=0.8051+ ( 0.4977∗20000 )+ ( 0.4733∗10 )
y=9959.54
Hence, number of mobile sold in a day would be 9960.
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