What is the approximate width of a 90 percent confidence interval for the true population proportion if there are 12 successes in a sample of 25?

To find for the value of the approximate width or margin of error, we make use of the following formula:

approximate width = ± z [p (1 - p) / n]^0.5

where,

z = z score can be obtained using normal distribution tables at 90% confidence level

p = probability of success

n = number of samples = 25

Using the standard distribution tables, the value of z at 90% level is:

z = 1.645

The probability of success is 12 out of 25, therefore:

p = 12 / 25 = 0.48

Therefore the approximate width is:

approximate width = ± 1.645 [0.48 (1 - 0.48) / 25]^0.5

approximate width = ± 1.645 (0.0999)

approximate width = ± 0.164