Regression Analysis - James McWhinney
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PART ONE: Problem Statement: Mr. James McWhinney, president of Daniel-James Financial Services, believe there is a relationship between the number of client contacts and the dollar amount of sales. To document this assertion, Mr. McWhinney gathered the following sample information. The X column indicates the number of client contacts last month, and the Y column shows the value of sales ($ thousand) last month for each client sampled. [Please note: You may use Excel for all of your calculations, however, accurate interpretation for each part is required]
Number of Sales
Contacts, ($ thousands)
X Y
14 24
12 14
20 28
16 30
46 80
23 30
48 90
50 85
55 120
50 110
a) Let the sales amount be the response (dependent variable ), and the number of contacts be the exploratory (independent variable ). Create a scatter diagram, using Excel/MegaStat.
b) Determine the coefficient of correlation ( ). Determine the coefficient of determination ( ). What proportion of total variation in sales can be explained or accounted for, by the variation in number of contacts. Use .05
c) Can it be concluded that there is positive correlation between the number of contacts and sales? (Use )
Step 1: State the Null and Alternative Hypotheses, symbolically,
using the population coefficient of correlation ( -
Pronounced "rho")
Step 2: State the Decision Rule: Find the critical value of the test,
using Student's t-distribution ( Appendix F), and state the
rejection region.
Step 3: Compute the test statistic, and estimate the p-value.
Step 4: Make Decision regarding the Null Hypothesis.
Step 5: Interpret the results:
d) Determine the regression equation. Interpret the slope value.
e) Estimate the sales amount, if there were 48 contacts.
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Solution Summary
The solution uses hypothesis testing and regression analysis to determine a regression equation for James McWhinney.
Education
- BSc , Wuhan Univ. China
- MA, Shandong Univ.
Recent Feedback
- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
- "excellent work"
- "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
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