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Data Interview Question

Advertisement Effectiveness

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Solution & Explanation

Objective: The task is to determine if the red advertisement is more effective than the blue advertisement based on the number of clicks each received.

Approach: To solve this, we will use a two-proportion Z-test. This statistical test is used to compare the proportions of two independent groups to see if there is a significant difference between them.

Step-by-Step Solution

1. Define the Hypotheses:

  • Null Hypothesis (H0H_0): The click-through rates for the red and blue ads are the same (pred=pbluep_{\text{red}} = p_{\text{blue}}).
  • Alternative Hypothesis (HaH_a): The click-through rate for the red ad is higher than the blue ad (pred>pbluep_{\text{red}} > p_{\text{blue}}).

2. Sample Proportions:

  • Blue Ad:
    • Clicks: 30
    • Total Shown: 100
    • Proportion (p^blue\hat{p}_{\text{blue}}): 30100=0.30\frac{30}{100} = 0.30
  • Red Ad:
    • Clicks: 40
    • Total Shown: 100
    • Proportion (p^red\hat{p}_{\text{red}}): 40100=0.40\frac{40}{100} = 0.40

3. Pooled Proportion:

  • The pooled proportion is calculated under the assumption that the null hypothesis is true (i.e., the proportions are equal). p^pooled=xblue+xrednblue+nred=30+40100+100=0.35\hat{p}_{\text{pooled}} = \frac{x_{\text{blue}} + x_{\text{red}}}{n_{\text{blue}} + n_{\text{red}}} = \frac{30 + 40}{100 + 100} = 0.35

4. Calculate the Standard Error (SE):

  • The standard error of the difference in proportions is: SE=p^pooled(1p^pooled)(1nblue+1nred)SE = \sqrt{\hat{p}_{\text{pooled}} \cdot (1 - \hat{p}_{\text{pooled}}) \cdot \left(\frac{1}{n_{\text{blue}}} + \frac{1}{n_{\text{red}}}\right)} SE=0.350.65(1100+1100)=0.004550.0675SE = \sqrt{0.35 \cdot 0.65 \cdot \left(\frac{1}{100} + \frac{1}{100}\right)} = \sqrt{0.00455} \approx 0.0675

5. Calculate the Z-Statistic:

  • The Z-statistic measures how many standard errors the observed difference is from 0: z=p^redp^blueSE=0.400.300.06751.48z = \frac{\hat{p}_{\text{red}} - \hat{p}_{\text{blue}}}{SE} = \frac{0.40 - 0.30}{0.0675} \approx 1.48

6. Determine the P-Value:

  • For a one-tailed test, the p-value is: p=P(Z>1.48)p = P(Z > 1.48)
  • Using a z-table or statistical software: p0.069p \approx 0.069

7. Conclusion:

  • At a significance level of α=0.05\alpha = 0.05:
    • Since p=0.069>0.05p = 0.069 > 0.05, we fail to reject the null hypothesis.
    • There is insufficient evidence to conclude that the red ad is significantly more effective than the blue ad.

Explanation

This test helps assess whether the observed difference in click-through rates between the red and blue ads is statistically significant. While the red ad had a higher click-through rate, the evidence is not strong enough to conclude that the difference is not due to random chance. Increasing the sample size or conducting further tests could provide more definitive results.