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…

Question

asked Jan 16, 2022 179k views

1 Answer

Namezero · Answer 1 · 2022-01-22T02:38:29+0000

Answer:

D. 77

Explanation:

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

$\alpha = (1 - 0.98)/(2) = 0.01$

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

$M = z(\sigma)/(√(n))$

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

$M = z(\sigma)/(√(n))$

80 = 2.327(300)/(√(n))

80√(n) = 2.327*300

√(n) = (2.327*300)/(80)

(√(n))^2 = ((2.327*300)/(80))^2

n = 76.15

Rounding up:

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