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) 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

fwms@lil_ben06

step-by-step explanation:

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v