Question

Switching banks after a merger. Banks that merge with others to form “mega-banks” sometimes leave customers dissatisfied with the impersonal service. A poll by the Gallup Organization found 20% of retail customers switched banks after their banks merged with another. One year after the acquisition of First Fidelity by First Union, a random sample of 250 retail customers who had banked with First Fidelity were questioned. Let$\widehat{\mathbf{p}}$ be the proportion of those customers who switched their business from First Union to a different bank.

1. Find the mean and the standard deviation of role="math" localid="1658320788143" $\widehat{\mathbf{p}}$.
2. Calculate the interval $\mathbf{E}\left(\widehat{p}\right)\mathbf{\pm }\mathbf{2}{\mathbf{\sigma }}_{\widehat{p}}$.
3. If samples of size 250 were drawn repeatedly a large number of times and determined for each sample, what proportion of the values would fall within the interval you calculated in part c?

Short answer

1. The mean and standard deviation of the proportion are 0.2and 0.0253 respectively.
2. The interval is (0.1494, 0.2506).
3. The proportion of the values that would fall within the interval of part b is 0.9544.

Step-by-step solution

Given information

A Gallup Organization study showed that 20% of retail customers switched banks after their banks merged with another. After one year, a random sample of 250 retail customers was questioned if they changed their business from the First union to a different bank.

Let$\widehat{p}$ be the proportion of those customers who switched their business from First Union to a different bank.

Determine the mean and variance

The mean of the proportion is equal to the true binomial proportion p. That is,

$\begin{array}{c}E\left(\widehat{p}\right)={\mu }_{\widehat{p}}\\ =p\\ =0.2\end{array}$

And the standard deviation of the proportion is defined as,

role="math" $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.2\left(1-0.2\right)}{250}}\\ =\sqrt{\frac{0.16}{250}}\\ =\sqrt{0.00064}\\ =0.0253\end{array}$

Thus, the mean and standard deviation of the proportion are 0.2and 0.0253 respectively.

Determine the interval

Consider the interval$E\left(\widehat{p}\right)\pm 2{\sigma }_{\widehat{p}}$

As there is obtained that$E\left(\widehat{p}\right)=0.2and{\sigma }_{\widehat{p}}=0.0253$

So, the interval is,

$\begin{array}{c}E\left(\widehat{p}\right)\pm 2{\sigma }_{\widehat{p}}=0.2\pm 2\left(0.0253\right)\\ =0.2\pm 0.0506\\ =\left(0.1494,0.2506\right)\end{array}$

Thus, the interval is (0.1494,0.2506).

Determine the proportion

Sample of size 250 were drawn repeatedly at a large number of times and determined the value of$\widehat{p}$ for each time.

In part b. there considered the interval (0.1494, 0.2506)

So, consider,

$\begin{array}{c}\mathrm{Pr}\left(0.1494\le \widehat{p}\le 0.2506\right)=\mathrm{Pr}\left(\frac{0.1494-p}{\sqrt{{\sigma }_{\widehat{p}}}}\le \frac{\widehat{p}-p}{\sqrt{{\sigma }_{\widehat{p}}}}\le \frac{0.2506-p}{\sqrt{{\sigma }_{\widehat{p}}}}\right)\\ =\mathrm{Pr}\left(\frac{0.1494-0.2}{0.0253}\le z\le \frac{0.2506-0.2}{0.0253}\right)\\ =\mathrm{Pr}\left(\frac{-0.0506}{0.0253}\le z\le \frac{0.0506}{0.0253}\right)\\ =\mathrm{Pr}\left(-2\le z\le 2\right)\\ =\mathrm{Pr}\left(z\le 2\right)-\mathrm{Pr}\left(z\le -2\right)\\ =0.5+0.4772-0.5+0.4772\\ =0.4772+0.4772\\ =0.9544\end{array}$

Therefore, the required proportion of the$\widehat{p}$ values that would fall within the interval of b is 0.9544.