Question

In 2008 the Better Business Bureau settled 75% of complaints they received (USA Today, March 2, 2009). Suppose you have been hired by the Better Business Bureau to investigate the complaints they received this year involving new car dealers. You plan to select a sample of new car dealer complaints to estimate the proportion of complaints the Better Business Bureau is able to settle. Assume the population proportion of complaints settled for new car dealers is .75, the same as the overall proportion of complaints settled in 2008

Suppose you select a sample of 450 complaints involving new car dealers. Show the sampling distribution of p¯.

Answer 1

`Mean=p=0.75`

`sd=\sqrt{(p(1-p))/(n)}=\sqrt{(0.75(1-0.75))/(450)}=0.0204`

Explanation:

1) Data given

`p=75\%=0.75` represent the population proportion of complaints settled for new car dealers

`n=450` represent the sample of complaints involving new car dealers.

2) Find the distribution of
`\hat p`

First we can begin with the expected value

`E(\hat p)=p` and that represent the mean

Now we can find the variance for
`\hat p`

When we use a proportion p, when we draw n items each from a Bernoulli distribution. The variance of each Xi distribution is p(1−p) and hence the standard error is p(1−p)/n. for this reason the variance for
`\hat p` is given by:

`Var(\hat p)= (p(1-p))/(n)`

So then the deviation would be given by:

`Sd(\hat p)=\sqrt{(p(1-p))/(n)}`

The sample distribution of the sample proportion
`\hat p` is normal, so then we have this:

`\hat p \sim N(p,\sqrt{(p(1-p))/(n)})`

3) Calculating the mean and standard deviation

We can replace the values given in order to find the mean and deviation:

`Mean=p=0.75`

`sd=\sqrt{(p(1-p))/(n)}=\sqrt{(0.75(1-0.75))/(450)}=0.0204`