Question

A company is considering two different campaigns, A and B, for the promotion of their product. Two tests are conducted in two market areas with identical consumer characteristics, and in a random sample of 60 customers who saw campaign A, 18 tried the product. In a random sample of 100 customers who saw campaign B, 22 tried the product. What conclusion can management reach?

Answer 1

Answer:

There is no sufficient evidence of a difference in the effectiveness of the two ad campaigns.  

Step-by-step explanation:

We are given the following in the question:

$x_1 = 18\\n_1 = 60\\x_2 = 22\\n_2 = 100$

Let p1 and p2 be the proportion from campaign A and B, respectively, that are defective.

$p_1 = \dfrac{x_1}{n_1} = \dfrac{18}{60} = 0.3\\\\p_2 = \dfrac{x_2}{n_2} = \dfrac{x_2}{n_2} = \dfrac{22}{100}= 0.22$

First, we design the null and the alternate hypothesis

$H_{0}: p_1 = p_2\\H_A: p_1 \neq p_2$

We use Two-tailed z test to perform this hypothesis.

Formula:

$\text{Pooled P} = \dfrac{x_1+x_2}{n_1+n_2}\\\\Q = 1 - P\\\\Z_{stat} = \dfrac{p_1-p_2}{\sqrt{PQ(\frac{1}{n_1} + \frac{1}{n_2})}}$

Putting all the values, we get,

$\text{Pooled P} = \dfrac{18+22}{60+100} = 0.25\\\\Q = 1 - 0.25 = 0.75\\\\Z_{stat} = \dfrac{0.3-0.22}{\sqrt{0.25\times 0.75(\frac{1}{60} + \frac{1}{100})}} = 1.1313$

Now, we calculate the p-value from the table at 0.05 significance level.

P-value = 0.25789

Since the p-value is greater than the significance level, we fail to reject the null hypothesis and accept it.

Conclusion:

Thus, we conclude that there is no sufficient evidence of a difference in the effectiveness of the two ad campaigns.