Awell-known bank credit card firm wishes to estimate the proportion of credit card holders who carry a nonzero balance at the end of the month and incur an interest charge. assume that the desire margin of error is 0.03 at 98% confidence.
a) how large a sample should be selected if it is anticipated that roughly 70% of the firm's card holders carry a nonzero balance at the end of the month?
b) how large a sample should be selected if no planning value for the proportion could be specified?

a) 1267

b)  1509

Step-by-step explanation:

As per given , we have

Significance level :

Using the z-value table , the critical z value for 98% confidence :

Margin of error : E= 0.03

Formula for sample size :

, where p is the prior estimate of population proportion.

a) p=0.70

Sample size :

Simplify , we get

Hence, the minimum sample size required = 1267

a) if no estimate of prior population proportion is given , we take p= 0.5

Sample size :

Simplify , we get

Hence, the minimum sample size required = 1509

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!

part 1 answer = 8

part 2 answer = 72

step-by-step explanation:

part 1 explanation = add all of the numbers, then divide by how many numbers you have. so, 200/25 = 8.

part 2 explanation = not gonna lie, this is hard to explain, so just google how to do it.