A Bank in Boston has 700 checking account customers?A Bank in Boston has 700 checking account customers. A sampoe of 64 of these customers showed 48 also had their home mortgages with the bank. Construct the 90 percnet confidence interval for the proportion of checking account customers who have a mortgage with the bank. | |
Since the relatively large (greater than 30) sample size is known but neither the standard deviation nor the mean are known, the central limit theorem can be invoked and the confidence interval for a population proportion must be determined. The confidence interval for a population proportion is given by the following:
p_hat +/- (z_a/2) * sqrt[(p_hat * q_hat)/n] = p_hat +/- (z_a/2) * sqrt[(p_hat * (1 - p_hat))/n].
The objective of this problem is to determine the confidence interval for a percentage of people that have a home mortgage with the bank that is proportional to a large population.
To help with understanding the general idea of using such a confidence interval, consider the process of distributing a survey. Suppose that you are wanting to distribute a survey and you want everyone in the entire world to fill-out it. But then you realize that is nearly impossible. You then need to find a way to take a sample n of a large population or the entire population and somehow find a way to let a small sample of size n be an indicator and/or estimator of the large population or the population at whole, to which it must be proportional. You then realize that you can apply information given by a confidence interval to estimate the measurements or results of the survey of the smaller sample size as compared to the entire population. When you are working with a sample of a larger population, there is always going to be some margin of error; margin of error = (z_a/2) * sqrt[(p_hat * q_hat)/n]. Assuming that the margin of error is small, the measurements or surveys that fall within this margin of error can be neglected. Margins of error that are small correspond to the confidence level that is chosen.
It is given that a sample size n of the population of 700 checking account customers is equal to 64. Out of these 64 customers, 48 had an additional home mortgage; hence, the percentage p_hat of the sample size is equal to 48 / 64 = 0.75.
Since P(-z <= Z <= z) ~ 1 - a = 0.90 (the given confidence level) implies that a = 0.10, so P(Z<=z) = phi(z_0.10/2) = 1 - 0.10/2 = 1 - 0.05 = 0.95. Then z_0.10/2 = phi^-1(z_0.10/2) = phi^-1(1 - 0.05) = phi^-1(0.95) = 1.64.
Now, p_hat +/- (z_a/2) * sqrt[(p_hat * q_hat) / n] = p_hat +/- (z_a / 2) * sqrt[(p_hat * (1 - p_hat)) / n] = 0.75 +/- (1.64) * sqrt[(0.75 * 0.25) / 64] = 0.75 +/- 0.089 or (0.66,0.84). Now, there is confidence in concluding that, between 66% and 84% of checking account customers have an additional mortgage with the bank.
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