The customer service manager for the XYZ Fastener Manufacturing Company examined 60 vouchers and found 9 vouchers containing errors. Find a 98 percent confidence interval for the proportion of vouchers with errors.

= (0.043, 0.257)

Explanation:

p = 9/60 = 0.15

Z score for 98% confidence interval = Z0.01 = 2.33

The Confidence interval = (p + Z0.01 * sqrt (p * (1 - p) / n))

= (0.15 + 2.33 * sqrt (0.15 * (1 - 0.15) / 60))

= (0.15 + 0.107)

= (0.043, 0.257)

The Confidence Interval (CI) is calculated as (0.043, 0.257).

The steps and expanation is shown below.

Explanation:

The formula for a Confidence Interval (CI) for a population proportion is given as

p + z*[Sqrt (p (1 - p)) / n]

Or

p - z*[Sqrt (p (1 - p)) / n],

p is the sample proportion, n is the sample size, and z * is the appropriate value from the standard normal distribution for your desired confidence level. The following table shows values of z * for certain confidence levels.

For the 98% Confidence Interval, z*-value = 2.33

p = 9/60 = 0.15, confidence level z * = 2.33 (From the standard table)

0.15 + 2.33[Sqrt (0.15 (1 - 0.15)) / 60]

OR

0.15 - 2.33[Sqrt (0.15 (1 - 0.15)) / 60]

0.15 + 0.107 or 0.15 - 0.107

Confidence Interval is

(0.043, 0.257)