Respuesta :
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:
[tex]x_1 = 18\\n_1 = 60\\x_2 = 22\\n_2 = 100[/tex]
Let p1 and p2 be the proportion from campaign A and B, respectively, that are defective.
[tex]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[/tex]
First, we design the null and the alternate hypothesis
[tex]H_{0}: p_1 = p_2\\H_A: p_1 \neq p_2[/tex]
We use Two-tailed z test to perform this hypothesis.
Formula:
[tex]\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})}}[/tex]
Putting all the values, we get,
[tex]\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[/tex]
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.
