Need Stats help

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 128. What is the probability that the sample proportion will be at least 2 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.)

Answer 1

p-hat = sample proportion

p = population proportion, defined in the problem as p = 0.75

n = 128

The problem asks us to find the probability that the sample proportion (p-hat) is at least 2 percent higher than the population proportion (p = 0.75), so we want to find the probability that p-hat is greater than 0.77:

P(p-hat > 0.77)

To find a z-score for this type of problem, we use z = (p-hat - p)/std error, so next we need to calculate the standard error

The formula for the standard error of a mean proportion:

std error = sqrt[(p)*(1-p)/n], so in this case:

std error = sqrt(0.75*0.25/128) = 0.0383

Now we plug that into our z-score formula:

z = (0.77 - 0.75)/0.0383 = 0.5526

We want the probability that p-hat is GREATER than 0.77:

P(p-hat > 0.77) = P(z > 0.5526)

The z-table is easier to use for the left tail, so instead we do:

P(p-hat > 0.77) = P(z < -0.5526) = 0.2903

Answer 2

Given the information about this BBB survey, including:

p = 0.75; n = 128; p-hat = 0.77 (which comes from simply adding the 2% increase from the population proportion, p)

We'll apply Central Limit Theorem (CLT) to this sampling distribution of sample proportion (p-hat) to which is underpinned by the binomial probability distribution conditioning. (If you like - or maybe you already are aware of this source - please check out this page from this textbook link by OpenStax that goes into some very helpful details further: https://openstax.org/books/introductory-business-statistics/pages/7-3-the-central-limit-theorem-for-proportions). With that, we'll find the z-score of the p-hat value and therefore the probability of at least 0.77, or 77% of complaints were settled in a given sample.

This means we would need to identify the mean and standard error of this sampling distribution given p and n. You could also just plug all this into the CLT formula for sample proportions, as shown:

z = (p-hat - p) / sqrt[ (p*(1-p))/n ]

where p represents the "mean of the sampling distribution of p-hat" and the denominator -- square root of p times 1-p divided by n -- is the "standard error of the sampling distribution of p-hat".

So, seen separately, we have:

mean = p = 0.75

standard error (SE) = sqrt[ 0.75*(1-0.75) / 128 ] = 0.0382732772 (or approx. 0.03827)

Altogether there, we get z = (0.77 - 0.75) / 0.0382732772 = 0.5226 approximately

Now, with your class, it seems, using this "wamap approach", I will leave it to you on how exactly you find/calculate the probability from the z-score. I will mention a couple other ways that might relate to wamap. With our z-score there, you can take it to a Standard Normal Distribution chart where you cross-point the 0.0 places and .00 hundredth's place of the z-value we found to get the probability to the left (since most if not all z-charts are defaulted this way) and then subtract from 1 to get the probability to the right to represent this "at least" case, like so:

P(p-hat > 0.77) = P(z > 0.5226) = 1 - P(z < 0.5226)

However, you can also use some kind of calculator/technology that allows you to set it up easily to return the probability to the right of the z-score being 0.5226 - this is how I did it, using my TI-84 graphing calculator, which resulted in the probability being:

P(z > 0.5226) = 0.3006 approximately

I hope this was helpful to you in setting up and relating it to some important, fundamental concepts. :) Peace!