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 percent confidence interval is used and an interval of $75 is desired, how many cardholders should be sampled?a. 212
b. 87
c. 629
d. 44

B. 87

Step-by-step explanation:

The first thing is to calculate critical z factor

the alpha and the critical z score for a confidence level of 90% is calculated as follows:

two sided alpha = (100% - 98%) / 200 = 0.01

critical z factor for two sided alpha of .01 is calculated as follows:

critical z factor = z factor for (1 - .01) = z factor for (.99) which through the attached graph becomes:

critical z factor = 2.33

Now we have the following formula:

ME = z * (sd / sqrt (N) ^ (1/2))

where ME is the margin of error and is equal to 75, sd is the standard deviation which is 300 and the value of z is 2.33

N the sample size and we want to know it, replacing:

75 = 2.33 * (300 / (N) ^ (1/2))

solving for N we have:

N = (2.33 * 300/75) ^ 2

N = 86.86

Which means that the sample size was 87.

answer: 298143

step-by-step explanation:

i think 1: 2

step-by-step explanation: