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The null hypothesis (\(H_0\)) is that the success rate of the campaign is the same as the standard rate (i.e., \(p = 0.02\)), where p is the proportion of success. The alternative hypothesis (\(H_a\)) is that the success rate is different from the standard rate (i.e., \(p ≠ 0.02\)).

For conducting inference about a proportion, we must check if the conditions of Randomness, Normality, and Independence are met. Randomness condition seems to be met as the problem states that the offers were sent to a random sample. Normality condition can be checked using the large counts condition which requires both \(np ≥ 10\) and \(n(1-p) ≥ 10\). Here, \(n = 50000\) and \(p = 0.02\). Thus, both \(np = 50000*0.02 = 1000 ≥ 10\) and \(n(1-p) = 50000*0.98 = 49000 ≥ 10\), so the Normality condition is met. Independence condition can be assumed if the sample is less than 10% of population, as we usually don't have access to accurate population sizes, especially in this case, it's safe to assume that 50000 is less than 10% of all cardholders. Thus, all conditions for inference are satisfied.

As of now, based on the provided sample proportion \(2.4%\) which is slightly higher than the success rate they are used to seeing in similar campaigns \(2% = 0.02\), it might be tempting to think that the strategy is working. However, without performing a proper hypothesis test using a significance level (say, \(α = 0.05\)), it is premature to conclude that they will benefit from using this campaign in terms of the success rate. The prediction would require calculation of the test statistic and p-value using statistical software.