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

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.