Problem: Analysis of user purchasing behavior on e-commerce platforms
Background: An e-commerce platform wants to study whether the average purchase amount of users is affected by membership level (ordinary members vs. gold members). The most recent order amount (in US dollars) of 30 ordinary members and 30 gold members was randomly selected, and the data is as follows:
Ordinary members: sample mean = $85, sample standard deviation = $15
Gold members: sample mean = $110, sample standard deviation = $20
Task:
1. Descriptive statistics: Calculate the 95% confidence interval of the two sets of data and explain the results.
2. Hypothesis test: At the significance level α=0.05, test whether there is a significant difference in the average purchase amount of the two types of members.
Propose the null hypothesis (H₀) and the alternative hypothesis (H₁).
Choose an appropriate test method (such as independent sample t-test) and explain the reason.
Calculate the test statistic and draw a conclusion.
3. Additional thinking: If the p-value in the actual analysis is 0.03, but the platform manager explains that “the difference may be caused by sampling error”, how to respond?
Struggling with where to start this assignment? Follow this guide to tackle your assignment easily!
Introduction
Analyzing user purchasing behavior on e-commerce platforms is essential for understanding how different factors, such as membership levels, influence spending habits. This guide will walk you through the process of structuring and writing your paper, focusing on the analysis of average purchase amounts between ordinary members and gold members.
1. Introduction
Begin your paper with an introduction that outlines the purpose of the study. Clearly state that the objective is to determine whether membership level affects the average purchase amount on the e-commerce platform.
2. Descriptive Statistics
In this section, you’ll calculate and interpret the 95% confidence intervals for both groups.
a. Understanding Confidence Intervals
A confidence interval provides a range within which we can be certain, to a specific probability, that the population parameter lies. A 95% confidence interval means that if we were to take 100 different samples and compute a confidence interval for each sample, we would expect about 95 of the intervals to contain the true mean purchase amount.
b. Calculating the Confidence Intervals
Given data:
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Ordinary Members: Sample mean (Xˉ1\bar{X}_1Xˉ1) = $85, Sample standard deviation (s1s_1s1) = $15, Sample size (n1n_1n1) = 30
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Gold Members: Sample mean (Xˉ2\bar{X}_2Xˉ2) = $110, Sample standard deviation (s2s_2s2) = $20, Sample size (n2n_2n2) = 30
The formula for the confidence interval is:
Xˉ±t∗×sn\bar{X} \pm t^* \times \frac{s}{\sqrt{n}}Xˉ±t∗×nsStatistics LibreTexts
Where:
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Xˉ\bar{X}Xˉ = Sample mean
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t∗t^*t∗ = t-score corresponding to the desired confidence level and degrees of freedom
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sss = Sample standard deviation
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nnn = Sample size
For a 95% confidence level with 29 degrees of freedom (n−1=30−1n-1 = 30-1n−1=30−1), the t∗t^*t∗ value is approximately 2.045.PennState: Statistics Online Courses
Ordinary Members:
85±2.045×1530=85±5.6⇒[79.4,90.6]85 \pm 2.045 \times \frac{15}{\sqrt{30}} = 85 \pm 5.6 \Rightarrow [79.4, 90.6]85±2.045×3015=85±5.6⇒[79.4,90.6]
Gold Members:
110±2.045×2030=110±7.5⇒[102.5,117.5]110 \pm 2.045 \times \frac{20}{\sqrt{30}} = 110 \pm 7.5 \Rightarrow [102.5, 117.5]110±2.045×3020=110±7.5⇒[102.5,117.5]
c. Interpretation
The 95% confidence interval for ordinary members is between $79.4 and $90.6, and for gold members, it’s between $102.5 and $117.5. Since these intervals do not overlap, it suggests a potential difference in average purchase amounts between the two groups.
3. Hypothesis Testing
This section involves testing whether the observed difference in means is statistically significant.
a. Formulating Hypotheses
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Null Hypothesis (H0H_0H0): There is no difference in average purchase amounts between ordinary and gold members (μ1=μ2\mu_1 = \mu_2μ1=μ2).
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Alternative Hypothesis (H1H_1H1): There is a difference in average purchase amounts between ordinary and gold members (μ1≠μ2\mu_1 \neq \mu_2μ1=μ2).
b. Choosing the Test Method
An independent samples t-test is appropriate here because we are comparing the means of two independent groups to see if they differ significantly.
c. Calculating the Test Statistic
The formula for the t-statistic in an independent samples t-test is:
t=Xˉ1−Xˉ2s12n1+s22n2t = \frac{\bar{X}_1 – \bar{X}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}t=n1s12+n2s22Xˉ1−Xˉ2Statistics LibreTexts+2en.wikipedia.org+2PennState: Statistics Online Courses+2
Plugging in the values:
t=85−11015230+20230=−257.5+13.33=−2520.83=−254.56≈−5.48t = \frac{85 – 110}{\sqrt{\frac{15^2}{30} + \frac{20^2}{30}}} = \frac{-25}{\sqrt{7.5 + 13.33}} = \frac{-25}{\sqrt{20.83}} = \frac{-25}{4.56} \approx -5.48t=30152+3020285−110=7.5+13.33−25=20.83−25=4.56−25≈−5.48
d. Determining Degrees of Freedom
Using the Welch-Satterthwaite equation:en.wikipedia.org
df=(s12n1+s22n2)2(s12n1)2n1−1+(s22n2)2n2−1df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 – 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 – 1}}df=n1−1(n1s12)2+n2−1(n2s22)2(n1s12+n2s22)2en.wikipedia.org
Calculating:
df=(7.5+13.33)27.5229+13.33229=20.83256.2529+177.6929=433.698.06≈53.8df = \frac{(7.5 + 13.33)^2}{\frac{7.5^2}{29} + \frac{13.33^2}{29}} = \frac{20.83^2}{\frac{56.25}{29} + \frac{177.69}{29}} = \frac{433.69}{8.06} \approx 53.8df=297.52+2913.332(7.5+13.33)2=2956.25+29177.6920.832=8.06433.69≈53.8
Rounding down, we use 53 degrees of freedom.JMP Statistical Discovery
e. Making the Decision
With 53 degrees of freedom and a significance level (α\alphaα) of 0.05, the critical t-value for a two-tailed test is approximately ±2.005. Since our calculated t-value of -5.48 exceeds this in magnitude, we reject the null hypothesis, indicating a significant difference in average purchase amounts between the two membership levels.
4. Addressing the Manager’s Concern
If the p-value is 0.03, it indicates a statistically significant difference at the 0.05 level. However
