Answer:
The success rate for the full campaign will be different than the standard rate of 2%.
Step-by-step explanation:
To predict the results conduct a hypothesis test for single proportion.
Assume that the significance level of the test is [tex]\alpha =5\%\ or\ 0.05[/tex].
[tex]H_{0}:[/tex] The the success rate for the full campaign will be no different than the standard rate, i.e. p = 0.02
[tex]H_{a}:[/tex] The the success rate for the full campaign will be different than the standard rate, i.e. p ≠ 0.02
[tex]z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} }}[/tex]
Compute the value of the test statistic as follows:
[tex]z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} }}\\=\frac{0.024-0.02}{\sqrt{\frac{0.02(1-0.02)}{50000} }}\\=6.3887\\\approx 6.39[/tex]
The hypothesis test is two tailed. Then the for 5% level of significance the rejection region is defined as: [tex][-1.96\leq Z\leq 1.96][/tex], i.e if the test statistic value lies out of this region then the null hypothesis will be rejected.
The calculated value of the test statistic is z = 6.39.
That is, [tex]z=6.39>1.96[/tex].
Thus, we may reject the null hypothesis.
As the null hypothesis is rejected at 5% level of significance this implies that the success rate for the full campaign will be different than the standard rate of 2%.