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 276. What is the probability that the
sample proportion will be at least 5 percent more than the
population proportion?
Note: You should carefully round any z-values you calculate to at
least 4 decimal places to match wamap's approach and
calculations.
Answer = (Enter your answer as a number accurate to 4
decimal places.)
Solution
Given that,
p = 0.75
1 - p = 1 - 0.75 = 0.25
n = 276
_{} = p = 0.75
_{} = [p( 1 - p ) / n] = [(0.75 * 0.25) / 276 ] = 0.0261
= 0.75 + 0.05 = 0.80
P( > 0.80) = 1 - P( < 0.80)
= 1 - P(( - _{} ) / _{} < (0.80 - 0.75) / 0.0261 )
= 1 - P(z < 1.9157)
Using z table
= 1 - 0.9723
= 0.0277
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