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

56.6

step-by-step explanation:

the ratio of the perimeters is equal to the scale.

the ratio of the areas is equal to the square of the scale.

34 / 20 = k

a / 19.6 = k²

substituting:

a / 19.6 = (34 / 20)²

a = 19.6 (34 / 20)²

a ≈ 56.6

on monday rodney ate 1/4th of the sandwich = 3/12

on tuesday he ate 1/3rd of the sandwich = 4/12

therefore in total he has eaten = 3/12 + 4/12 = 7/12

thus we can say that he has eaten 7/12 of the sandwich on monday and tuesday alltogether.

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