Frequency and Percent Frequency Distributions of Furniture Order Values
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 presence of these higher values. However, this is not the case with median which does not take into consideration the exact values.
a.Whether demand and unit prices are related or not.
Alpha = 0.05
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
R Sqare ssr/sst=5048/8181=.617
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 determined under root 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.
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 u1=u2=u3
Alternative hypothesis H1 u1/=u2/=u3
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.
Number of mobile phones sold per day =y
Price of mobile phones =x1
Number of advertising spots =x2
a.Regression equation
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
Degree of freedom for regression = k = 2
Degree of residual
Now,
Null hypothesis B1=B2=B3….BK =0
Alternative hypothesis H1 at least bi is not eqal to 0
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
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.
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.
Number of advertising spots x2 = 10
Hence, number of mobile sold in a day would be 9960.
Order an Essay Now & Get These Features For Free:
Turnitin Report
Formatting
Title Page
Citation
Outline
