Question

The First National Bank of Wilson has 600 checking account customers. A recent sample of 50 of these customers showed 29 have a visa card with the bank. Construct the 90% confidence interval for the proportion of checking account customers who have a visa card with the bank.

Answer 1

Answer:

0.58 +/- 0.115

(0.465, 0.695)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

p+/-z√(p(1-p)/n)

Given that;

Proportion p = 29/50 = 0.58

Number of samples n = 50

Confidence interval = 90%

z(at 90% confidence) = 1.645

Substituting the values we have;

0.58 +/- 1.645√(0.58(1-0.58)/50)

0.58 +/- 1.645(0.069799713466)

0.58 +/- 0.114820528652

0.58 +/- 0.115

(0.465, 0.695)

The 90% confidence interval estimate of the true population proportion of account customers who have a visa card with the bank is;

0.58 +/- 0.115

(0.465, 0.695)