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?
To determine whether the superior performance of the new brochure is just due to chance or if it is actually better than the old brochure, we can conduct a hypothesis test. In this case, we want to compare the proportion of customers who call for more information between the two groups (standard brochure vs. new brochure).
We can set up the following hypotheses:
Null hypothesis (H0): The new brochure is no better than the old brochure. The proportion of customers who call for more information is the same for both groups.
Alternative hypothesis (Ha): The new brochure is superior to the old brochure. The proportion of customers who call for more information is higher for the new brochure group.
Now, let's follow these steps to conduct the hypothesis test:
Step 1: Define the level of significance (alpha)
In this case, the significance level is given as 0.10. This means that we are willing to accept a 10% chance of making a Type I error, which is rejecting the null hypothesis when it is true.
Step 2: Calculate the test statistic
To calculate the test statistic, we use the formula for the test of two proportions:
z = (p1 - p2) / sqrt(p * (1 - p) * (1/n1 + 1/n2))
Where:
- p1 is the proportion of customers calling for more information in the new brochure group
- p2 is the proportion of customers calling for more information in the standard brochure group
- p is the pooled proportion, calculated as (x1 + x2) / (n1 + n2)
- x1 is the number of customers calling for more information in the new brochure group
- x2 is the number of customers calling for more information in the standard brochure group
- n1 is the sample size of the new brochure group
- n2 is the sample size of the standard brochure group
Step 3: Determine the critical value(s)
Since this is a one-tailed test (we are testing if the new brochure is superior), we need to find the critical value for a one-tailed test with a significance level of 0.10. This critical value corresponds to the z-score that leaves a 10% area in the tail.
Step 4: Compare the test statistic with the critical value(s)
If the test statistic falls in the rejection region (beyond the critical value), we reject the null hypothesis. If the test statistic falls in the non-rejection region (within the critical value), we fail to reject the null hypothesis.
Step 5: Calculate the p-value
If the test statistic falls in the rejection region, we calculate the p-value. The p-value represents the probability of observing a test statistic as extreme as, or more extreme than, the one observed, assuming the null hypothesis is true. If the p-value is less than the significance level (alpha), we reject the null hypothesis. If the p-value is greater than alpha, we fail to reject the null hypothesis.
By following these steps, we can determine if it is possible that the superior performance of the new brochure was just due to chance.
Note: The values of x1, x2, n1, and n2 are not mentioned in the given information, so we cannot proceed with the calculation of the test statistic without this information.