Question

In a recent year, the Better Business Bureau settled 75% of complaints they received. (Source: USA Today, March 2, 2009) You have been hired by the Bureau to investigate complaints this year involving computer stores. You plan to select a random sample of complaints to estimate the proportion of complaints the Bureau is able to settle. Assume the population proportion of complaints settled for the computer stores is the 0.75, as mentioned above. Suppose your sample size is 92.

What is the probability that the sample proportion will be 4 or more percent below the population proportion? Answer ______ ( 4 decimal places.) Also, explain how did you find the z score at the end.

Answer #1

Solution

Given that,

p = 0.75

1 - p = 1 - 0.75 = 0.25

n = 92

_{}
= p = 0.75

_{}
= [p
( 1 - p ) / n] =
[(0.75 * 0.25) / 92 ] = 0.0451

0.75 + 0.04 = 0.79

P( < 0.79 )

= P[( -
_{}
) /
_{}
< (0.79 - 0.75) / 0.0451 ]

= P(z < 0.89)

Using z table,

= 0.8133

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