A procedure where a sample data is collected and statistically used for inferential purposes is known as hypothesis testing. There are two types of hypothesis null and alternative. Null hypothesis is a statement that support population mean in inferential statistics. Alternative hypothesis is a statement that complement null hypothesis. Inferential statistics enables statisticians to calculate the significance level of null hypothesis based on the data provided. There are several types of statistical test approach which derive ‘test statistics’. Test statistics obtained is used to make conclusion based on defined ‘decision rule’. Regression can also be applied to evaluate the nature of relationship between variable.
Question 1
a) Considering the model where only the price is used to predict the sales volume, test the model to determine if the model is valid.
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Develop research problem.
Sales for normal and luxury goods increases when price decrease. The higher the price the lower the demand and vice versa. Based on demand principle the research question would be; does the price influence sales of frozen desert pies?
Collect data.
| Sales | Price ($) |
| 380 | 7.5 |
| 430 | 4.5 |
| 470 | 6.4 |
| 450 | 7 |
| 490 | 5 |
| 340 | 7.2 |
| 300 | 7.9 |
| 440 | 5.9 |
| 450 | 5 |
| 300 | 7 |
State null and alternative hypothesis.
Null hypothesis; there is a relationship between price and sales
H O : ≠ 0
Alternative hypothesis; There is no relationship between price and sales
H 1 : 1 0
Define critical region value based on alpha used in your analysis.
Degree of confidence – 95%
Level of significance – 5% (α = 0.05)
If the probability obtained is less than 0.05 null hypothesis true.
Define decision rule.
α ≥ p-value > α
p-value ≤ α : if p-value is less than or equals to α
Fail to reject null hypothesis
α (0.05) < p-value : p-value equals greater than α
Reject null hypothesis
P-value obtained after analysis of data using statistical package is the actual probability of the sample data.
Select appropriate test method to apply based on nature of data collected
Tool to use – Independent sample t-test two tailed
Conduct the test to obtain ‘test statistics’.
t-Test: Independent Samples- Computed in excel
| t-Test: Independent Samples | ||
| Price ($) (A) | Sales (B) | |
| Mean | 6.34 | 405.00 |
| Variance | 1.40 | 4916.67 |
| Observations | 10.00 | 10.00 |
| Hypothesized Mean Difference | 0.00 | |
| df | 9.00 | |
| t Stat | -17.98 | |
| P(T<=t) one-tail | 0.00 | |
| t Critical one-tail | 1.83 | |
| P(T<=t) two-tail | 0.00 | |
| t Critical two-tail | 2.26 | |
Compare test statistics to critical region to determine verdict of null
P-value < α
< 0.05
Inference we fail to reject null hypothesis
Question b. Using slope to determine significance
Hypothesis
Null hypothesis; there is a relationship between price and sales
H O : ≠ 0
Alternative hypothesis; There is no relationship between price and sales
H 1 : 1 0
Decision Rule
If calculated value is outside critical value range reject null hypothesis
Analysis approach
Regression – excel pack
Results
| Regression | |||||||
| Multiple R | 0.70 | ||||||
| R Square | 0.49 |
df |
SS |
MS |
F |
Sign-F |
|
| Adjusted Error | 0.42 | Regression |
1 |
6.11 |
6.11 |
7.58 |
0.02 |
| Standard Error | 0.90 | Residual |
8 |
6.45 |
0.81 |
||
| Observations | 10 | Total |
9 |
12.56 |
|||
| Coeff | Stand-Error | t Stat | P-value | Low-95% | Upp-95% | Low-95.0% | Upp-95.0% | |
| Intercept | 11.10 | 1.75 | 6.33 | 0.00 | 7.06 | 15.14 | 7.06 | 15.14 |
| Sales (B) | -0.01 | 0.0043 | -2.75 | 0.02 | -0.02 | 0.00 | -0.02 | 0.00 |
Inferences
SALES= 11.10 - (0.01*PRICE)
Test Statistics
| Sales Coeff | -0.01 |
| Stand-Error | 0.00 |
| test statistics | -2.75 |
t = = -2.75
Critical Value Range
| α | 0.05 |
| α/2 | 0.025 |
| t1- α/2 | 0.975 |
| df | 18 |
| critical value | ±2.10 |
Conclusion
-2.75 is outside ±2.10 hence, we reject null hypothesis
P-value approach
| p-value approach | |
| computed p-value | 2.3212E-08 |
| probability (-2.75)= | 2.3212E-08 |
Probability 2P(-2.75) = 2.3 ( )
Conclusion
Price and other factors influence sales
Question c. Coefficient of Determination
| r | 0.70 | ||
| Coefficient of Determination | 0.49 | ||
Question d . standard error of the estimate
Standard error = 0.0043
Question 2 advertisement cost vs. sales regression analysis
| Regression Statistics | |||||||||
| Multiple R | 0.58 | ||||||||
| R Square | 0.34 | ||||||||
| Adjusted R Square | 0.26 | ||||||||
| Standard Error | 38.16 | ||||||||
| Observations | 10 | ||||||||
| ANOVA | |||||||||
| df | SS | MS | F | Significance F | |||||
| Regression | 1 | 6041.19 | 6041.19 | 4.15 | 0.08 | ||||
| Residual | 8 | 11648.81 | 1456.10 | ||||||
| Total | 9 | 17690.00 | |||||||
| Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | Lower 95.0% | Upper 95.0% | ||
| Intercept | 201.36 | 74.45 | 2.70 | 0.03 | 29.67 | 373.04 | 29.67 | 373.04 | |
| SALES | 0.37 | 0.18 | 2.04 | 0.08 | -0.05 | 0.79 | -0.05 | 0.79 | |
From the analysis
SALES EQUATION = 201.36+0.37*ADVERT COST
Comparison
Sales predicting equation on price
SALES= 11.10 - (0.01*price)
From above equation and actual sales advertisement rate is a better predictor than price
Analyzed example
| Week | Sales | Price ($) | Advertising ($100s) |
| 1 | 380 | 7.5 | 4 |
Price analysis = 11.10 - (0.01*price)
| PRICE |
7.5 |
|||
| constant |
11.1 |
|||
| coeff | (-0.01) | |||
| SALES |
11.025 |
|||
| Advertisement Prediction | ||||
| advert cost | 400 | |||
| constant | 201.36 | |||
| coeff | 0.37 | |||
| SALES | 349.36 | |||
Reference
11.5 Regression. (nd). Retrieved from: http://uregina.ca/~gingrich/regr.pdf
SAGE. (2017). Comparing Two Groups Mean: The Independent Sample t Test [pdf]. Retrieved from:
http://oak.ucc.nau.edu/rh232/courses/EPS525/Handouts/Understanding%20the%20Independe nt%20t%20Test.pdf
