Only the formula, Thank you.

A bank manager has been presented with a new brochure that was designed to be more effective in attracting current customers to a personal financial counseling session that would include an analysis of additional baking services that could be advantageous to both the bank and the customer. The manager's assistant, who created the new brochure, randomly selects 400 current customers, then randomly chooses 200 to receive the standard brochure that has been used in the past, with the other 200 receiving the promising new brochure that he has developed. Of those receiving the standard brochure, 35% call for more information about the counseling session, while 42% of those receiving the new brochure call for more information. Using the 0.10 level of significance, is it possible that the superior performance of the new brochure was just due to chance and that the new brochure might really be no better than the old ones?

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To determine whether the superior performance of the new brochure is just due to chance or if it is significantly better than the old brochure, we can conduct a hypothesis test.

Let's define our null hypothesis (H0) and alternative hypothesis (H1):

H0: The new brochure is no better than the old brochure (p1 = p2)
H1: The new brochure is significantly better than the old brochure (p1 > p2)

Where p1 represents the proportion of customers who call for more information after receiving the new brochure, and p2 represents the proportion of customers who call for more information after receiving the old brochure.

Next, we need to calculate the test statistic and the critical value. In this case, we will use the z-test since we have the sample proportions and sample sizes.

The test statistic (z) is calculated using the formula:
z = (p1 - p2) / sqrt((p * (1 - p)) * ((1 / n1) + (1 / n2)))

Here, p is the pooled proportion, calculated as (x1 + x2) / (n1 + n2), where x1 and x2 are the number of customers who called for more information from the new and old brochures respectively, and n1 and n2 are the sample sizes.

The critical value is obtained from the standard normal distribution table based on the desired level of significance, which is 0.10 in this case.

If the calculated test statistic (z) is greater than the critical value, we reject the null hypothesis and conclude that the new brochure is significantly better. If the test statistic is less than the critical value, we fail to reject the null hypothesis.

Finally, we compare the calculated p-value with the level of significance (0.10) to make a decision. If the p-value is less than or equal to the level of significance, we reject the null hypothesis.

Now you can use the given data and the formulas provided to calculate the test statistic (z) and the critical value, and then make a decision based on the comparison of the test statistic with the critical value and the p-value with the level of significance.