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)

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