A well-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?

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jlmpco jlmpco · Answer 1 · 2019-11-11T17:20:14+01:00

Answer:

a) 1263 b) 1503

Step-by-step explanation:

Sample size for an Interval estimate of population proportion is

n = (zα/2)^2 p (1-p) / E^2

Sample size for an Interval estimate of a population mean, p is unknowmn

n = (zα/2)^2 0.25 / E^2

given:

proportion, p = 70

margin of error, E = 0.03

Confidence level of 98%, that means the the siginficance level α is 1 – p

α = 1 – 0.98 = 0.02

Z(α /2) = Z(0.02/2) = Z (0.01)

Using a Z table Z = 2.326

a) n = (Zα/2)2 p (1-p) / E2

n = 2.326^2*0.7 (1-.7)/0.03^2

n = 1262.39 = 1263

b) n = (Zα/2)2 0.25 / E2

n = 2.326^2 *0.25/0.03^2

n = 1502.85 = 1503