Question

A bank wishes to estimate the mean balances owed by customers holding Mastercard. The population standard deviation is estimated to be $300. If a 98% confidence interval is used and the maximum allowable error is $80, how many cardholders should be sampled?
A. 76
B. 85
C. 86
D. 77

Answer 1

Answer:

D. 77

Step-by-step explanation:

We have to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.98}{2} = 0.01$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$.

That is z with a p-value of $1 - 0.01 = 0.99$, so Z = 2.327.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

The population standard deviation is estimated to be $300

This means that $\sigma = 300$

If a 98% confidence interval is used and the maximum allowable error is $80, how many cardholders should be sampled?

This is n for which M = 80. So

$M = z\frac{\sigma}{\sqrt{n}}$

$80 = 2.327\frac{300}{\sqrt{n}}$

$80\sqrt{n} = 2.327*300$

$\sqrt{n} = \frac{2.327*300}{80}$

$(\sqrt{n})^2 = (\frac{2.327*300}{80})^2$

$n = 76.15$

Rounding up:

77 cardholders should be sampled, and the correct answer is given by option d.