Surname 1 Name Course title Professor Date Assignment Question 1 In this case, the aim is to find out if there exists any relationship between age of customers and the number of orders the customers place in a given year. The sample is of size 25. Below is the null and alternative hypothesis that will be applied in testing for the relationship. H 0 :There is no any linear relationship that exist between the variables H a : There is a linear relationshi p existing between the variables . Model Summary Mode R l 1 .957a R Adjusted R Std. Error of Square Square the Estimate .916 .913 Change Statistics R Square F Change Change 3.062 .916 251.663 df1 df2 Sig. F Change 1 23 .000 a. Predictors: (Constant), Number of orders placed in a year From the above output, the R2=0.916. This value indicates clearly that there is a very strong relationship between the independent variable and the dependent variable. That is, there is a very strong linear relationship between the age of customers and the number of orders customers' place in a given year. The p-value < 0.05 hence we will reject the null hypothesis and Surname 2 accept the claim hence conclude that there is a strong linear relationship existing between the variables at 5% level of significance. Coefficientsa Model Unstandardized Coefficients Standardized t Sig. Coefficients B (Constant) 1 Number of orders placed in a year Std. Error 26.192 1.619 2.751 .173 Beta .957 16.182 .000 15.864 .000 a. Dependent Variable: Age of customers The model will be, Age of customer=26.192+2.751 number of orders 1) What is the expected number of orders placed by a 55 year old customer? Age of customer=26.192+2.751 number of orders Using the above model, the number of orders will be, 55=26.192+2.751 number of orders number of orders=10.5 11 orders Question 2 In this case, the main objective is to find out if there exists a relationship between the rice of steak (per pound) and the total sales of steak in a week. The sample size is of 15 weeks. Below is the null and alternative hypothesis for testing the claim. H 0 :There is no any linear relationship that exist between the variables H a : Thereis a linear relationshi p existing between the variables . Surname 3 Model Summaryb Model R R Adjusted R Std. Error of Square Square the Estimate .953a 1 .909 .901 Change Statistics R Square F Change Change .30258 .909 df1 df2 Sig. F Change 129.103 1 13 .000 a. Predictors: (Constant), Weekly sales expressed as thousands of dollars b. Dependent Variable: Price per pound of steak in dollars From the above output, the R2=0.909. This is a clear indication that there is a very strong relationship between the price of steak and the total sales of steak in a week. The p-value < 0.05 hence we will reject the null hypothesis and conclude that there is a strong linear relationship existing between the price of steak and the total sales of steak in a week at 5% level of significance. Coefficientsa Model Unstandardized Coefficients Standardized t Sig. Coefficients B (Constant) 1 Weekly sales expressed as thousands of dollars Std. Error 7.896 .337 -.120 .011 Beta -.953 23.454 .000 -11.362 .000 a. Dependent Variable: Price per pound of steak in dollars The model for this is, Price of steak=7.8960.120 weekly sales . 1. What is the expected weekly sales when the company charges $3.75 for a pound of steak? Price of steak=7.8960.120 weekly sales . Surname 4 Using the above model, the weekly sales will be, $ 3.75=7.8960.120 weekly sales . weekly sales=$ 34.55 Question 3 In this scenario, we are interested in finding out the impact of advertising and coupon on sales. The sample size is of 12 months. Below is the null and alternative hypothesis for testing the claim. H 0 :There is no any linear relationship that exist between the variables H a : Thereis a linear relationshi p existing between the variables . Model Summary Model 1 R R Square .920a Adjusted R Std. Error of the Square Estimate .847 .813 21.36809 a. Predictors: (Constant), Dollar value of coupon, Number of Ads From the above output, the R2=0.847. This indicates that there is a very strong relationship between the sales made and advertisements and coupon. The p-value < 0.05 hence we will reject the null hypothesis and conclude that there is a strong linear relationship existing between the sales made and advertisements and coupon at 5% level of significance. Coefficientsa Model Unstandardized Coefficients Standardized t Sig. Coefficients B (Constant) 1 Number of Ads Dollar value of coupon Std. Error -52.081 20.651 7.642 1.310 28.912 5.534 a. Dependent Variable: Sales in thousands of dollars Beta -2.522 .033 .781 5.832 .000 .699 5.224 .001 Surname 5 The model for this is, Sales=7.642 ads+28.912 coupon52.081 1. what are the expected sales in a month where 8 advertisements were put out using a $3 coupon Sales=7.642 ads+28.912 coupon52.081 Using the above model, the sales will be, Sales=7.642(8)+28.912(3)52.081 Sales=$ 37.967 Question 4 In this case, we are interested in investigating the impact of price and advertising on beer sales. The sample size is of 12. Below is the null and alternative hypothesis for testing the claim. H 0 :There is no any linear relationship that exist between the variables H a : Thereis a linear relationshi p existing between the variables . Model Summary Model 1 R .999a R Adjusted R Std. Error of Square Square the Estimate .998 .998 Change Statistics R Square F Change Change 2.09637 a. Predictors: (Constant), Number of Ads Placed, Price in dollars .998 2776.859 df1 df2 Sig. F Change 2 9 .000 Surname 6 From the above table, the R2 = 0.998. This indicates that there is a very strong relationship between the dependent and independent variables involved. The p-value < 0.05 hence we will reject the null hypothesis and conclude that there is a strong linear relationship existing between the sales made and advertisements and price of beer at 5% level of significance. Coefficientsa Model Unstandardized Coefficients Standardized t Sig. Coefficients B 1 Std. Error (Constant) -4.796 3.221 Price in dollars -2.870 3.022 Number of Ads Placed 12.674 .175 Beta -1.489 .171 -.013 -.950 .367 1.002 72.457 .000 a. Dependent Variable: Sales in thousands of dollars The model for this is, Sales=12.674 ads2.87 price4.796 1. predict the beer sales when the company spends on 12 advertisements and prices the beer at $1.25 Sales=12.674 ads2.87 price4.796 Using the above model, the sales will be, Sales=12.674(12)2.87 (1.25)4.796 Sales=$ 143.7045