First USA, a major credit card company, is planning a new offer for their current cardholders. The offer will give double airline miles on purchases for the next 6 months if the cardholder goes online and registers for the offer. To test the effectiveness of the campaign, First USA recently sent out offers to a random sample of 50,000 cardholders. Of those, 1184 registered. Give a 95% confidence interval for the true proportion of those cardholders who will register for the offer.

Answer:

The answer is "[tex]0.02235 \ and \ 0.02501[/tex]".

Step-by-step explanation:

Given:

[tex]x= 1184\\\\n = 50000\\\\c = 95\% = 0.95[/tex]

The sample proportion is computed by subtracting that sample size by both the successes:

[tex]\hat{p}=\frac{x}{n} =\frac{1184}{50000}= 0.02368[/tex]

Determine [tex]z_{\frac{\alpha}{2}} = Z_{0.025}[/tex] using table Z in Table Z for trust level [tex]1 -\alpha = 0.95[/tex]

add using appendix F ((0.025) inside the table, therefore the z-score is the found z-score with negative sign):

So, the boundaries of the confidence interval:

[tex]\to \hat{p}-z_{\frac{\alpha}{2}} \cdot \sqrt{\frac{\hat{p} (1-\hat{p})}{n}} = 0.02368 - 1.96 \cdot \sqrt{\frac{0.02368(1-0.02368)}{50000}}\approx 0.02235 \\\\\to \hat{p}-z_{\frac{\alpha}{2}} \cdot \sqrt{\frac{\hat{p} (1-\hat{p})}{n}} = 0.02368 + 1.96 \cdot \sqrt{\frac{0.02368(1-0.02368)}{50000}}\approx 0.02501\\\\[/tex]

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