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)

being born in poverty would limit the people's chances, opportunities and benefits.

this happen because of several things, such as:

- poor people tend to gain less access to good education and nutrition. this mean that their chance to obtain high paying jobs would also be limited.

- poor people tend to socialize with other poor people while the rich tend to socialize with other rich people. this means that simply throguh their social circle, rich people would have more business opportunities compared to the poor people.

- rich people tend to have more capital in their disposal. this would benefit them in business since it is far easier for them to fund their business operation compared to poor people.

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