The Sample Proportion

Suppose a student guesses at the answer on every question in a 300-question exam. If he gets 60 questions correct, then his proportion of correct guesses is 60/300=.20. If he gets 75 questions correct, then his proportion of correct guesses is 75/300=.25. The proportion of correct guesses is simply the number of correct guesses divided by the total number of questions.

Similarly, if TV World sells 60 extended warranties with 300 television sets sold, then its warranty sales rate is 60/300=.20. If it sold 75 warranties, this is a sales rate of 75/300=.25. The warranty sales rate, or proportion of TV sets sold with warranties, is simply the number of warranties divided by the total number of TV sets.

Now, let X denote the number of successes out of a sample of n observations. If each observation is a success with probability p independently of the other observations, then X is a binomial random variable with parameters n and p. Furthermore, the proportion of successes  in the sample is also a random variable and is computed as 

$$\hat{p} = \frac{X}{n} = \frac{\mbox{number of successes}}{\mbox{total number of observations in the sample}}$$ (5.1)

Since X is expected to be around np give or take $\sqrt{np(1-p)}$, then X/n is expected to be around np/n give or take $\sqrt{np(1-p)}/n$, or p give or take $\sqrt{p(1-p)}/\sqrt{n}$. ( Make sure that you agree with the last statement before moving on. It may help to think of this analog: Suppose annual rainfall in Kalamazoo is expected to be around 24 inches give or take 6 inches. How do we change the measurement from inches to feet? We divide both numbers by 12! In feet, annual rainfall in Kalamazoo is expected to be around 24/12 give or take 6/12, or 2 feet give or take .5 feet. Now read the first sentence of this paragraph one more time.)

Going back to the TV World example, the number of warranties sold is expected to be around $60 \pm 7$. Thus the proportion of warranties sold is expected to be around $60/300 \pm 7/300$, or $.20 \pm .02$.

We summarize the formulas for the mean and SD of X and $\hat{p}$ in the following table .  

$$\begin{tabular}{ccc} \underline{Random Variable} & \underli... ...$\space & $p$\space & $\sqrt{p(1-p)}/\sqrt{n}$\\ \end{tabular}$$ (5.2)

Exercise 1. If TV World sold 1200 television sets last year,
a. the proportion of sets sold with extended warranties should be be around .20, give or take ________.
b. the percentage of sets sold with extended warranties should be around 20%, give or take ________.

Data analysis sometimes involves percentages  instead of proportions. The only difference, of course, is 2 decimal places, or multiplication by 100% (e.g. 20% instead of .20). In order to avoid repetition, we present all statistical formulas in proportions. As Exercise 1 shows, the answers can always be converted to percentages in the end.

Exercise 2. Historically, 5% of rentals from Campus Video are returned late.
a. Campus Video rented out 100 videos yesterday. The percentage that will be returned late should be around 5%, give or take ________.
b. Campus Video rented out 700 videos last week. The percentage that will be returned late should be around 5%, give or take ________. (Should the answer be different or the same as (a)?)

Exercise 2 is an illustration of the law of large numbers . A simpler illustration involves a coin toss. If you toss a coin repeatedly, which tends to get closer to 50% heads: 100 tosses or 700 tosses? The correct answer is 700. The larger the number of tosses, the closer we expect to get to 50%. The reason for this, as the Exercise shows, is the smaller give or take value. The sample size n lies in the denominator of the SD of $\hat{p}$. Therefore the larger the sample size, the smaller the SD, which happens to be the give-or-take value.