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 229. What is the probability that the sample proportion will be within 10 percent of the population proportion?

Answer 1

Answer:

99.96% probability that the sample proportion will be within 10 percent of the population proportion

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

Proportion p = 0.75

Mean:

$\mu = p = 0.75$

Standard deviation of the proportion:

$\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.75*0.25}{229}} = 0.0286$

What is the probability that the sample proportion will be within 10 percent of the population proportion?

This is the pvalue of Z when X = 0.75+0.1 = 0.85 subtracted by the pvalue of Z when X = 0.75 - 0.1 = 0.65. So

X = 0.85

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.85 - 0.75}{0.0286}$

$Z = 3.49$

$Z = 3.49$ has a pvalue of 0.9998

X = 0.65

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.65 - 0.75}{0.0286}$

$Z = -3.49$

$Z = -3.49$ has a pvalue of 0.0002

0.9998 - 0.0002 = 0.9996

99.96% probability that the sample proportion will be within 10 percent of the population proportion