Before we construct our distribution of the sample proportions, we first construct a distribution for the number of occurrences of successes of particular characteristic within our sample. Hence, we can let $X$X be the number of successes in our random sample. In addition to this, when the population is large or the sampling is done with replacement, then $X$X can be modelled by a binomial distribution such that $X\sim B\left(n,p\right)$X~B(n,p).
When we think about how we obtained the $\hat{p}$^p values in our example above, we know we simply divided each $x$x by $n$n, the possible number of outcomes, also known as the size of our sample.
We can relate this back to what we learnt about the linear change of scale and origin and we can therefore state that $\hat{P}=\frac{X}{n}$^P=Xn and hence:
$E\left(\hat{P}\right)$E(^P) | $=$= | $\frac{E\left(X\right)}{n}$E(X)n |
Using what we know about the linear change of scale |
$=$= | $\frac{n\times p}{n}$n×pn |
Substituting the expected value of $E\left(X\right)$E(X) formula from our binomial distribution |
|
$=$= | $p$p |
Simplifying |
Similarly we can apply the same linear change of scale and see how we calculate the standard deviation.
$SD\left(\hat{P}\right)$SD(^P) | $=$= | $\frac{\sqrt{n\times p\times\left(1-p\right)}}{n}$√n×p×(1−p)n |
Using what we know about the linear change of scale and our formula for the standard deviation of our binomial distribution |
$=$= | $\sqrt{\frac{n\times p\times\left(1-p\right)}{n^2}}$√n×p×(1−p)n2 |
Bringing the $n$n into the square root by squaring |
|
$=$= | $\sqrt{\frac{p\times\left(1-p\right)}{n}}$√p×(1−p)n |
Simplifying the $n$n and $n^2$n2 |
When selecting a random sample of size $n$n from a large population, the distribution of sample proportions has:
In the random sample of $5$5 students at a particular school, records show that $20%$20% of students walk to school.
(a) Calculate the mean of the distribution of the sample proportions.
Think: We could calculate the mean or expected value in the same way we did for any discrete random variable, and calculate directly from the table. However, since we know that the values in the table for $P\left(\hat{P}=\hat{p}\right)$P(^P=^p) are directly related to the values for $P\left(X=x\right)$P(X=x), which in turn are obtained from the binomial distribution, we can use what we know about calculating the expected value for the binomial distribution.
Do: From our work above we can see that $E\left(\hat{P}\right)=\hat{p}=p$E(^P)=^p=p which is $0.2$0.2 for this example.
(b) Calculate the standard deviation of the distribution of the sample proportions.
Think: Again, we could calculate the standard deviation directly from the table, or we can use the relationship we now know between the distribution of the sample proportions and the binomial distribution and get straight into our calculation.
Do: Using our simplified formula from above:
$SD\left(\hat{P}\right)$SD(^P) | $=$= | $\sqrt{\frac{0.2\times0.8}{5}}$√0.2×0.85 |
$=$= | $=\frac{2\sqrt{5}}{25}$=2√525 |
$15%$15% of all customers at a book store bought at least one autobiography. In a random sample of $100$100 customers, determine:
The expected number of customers who purchased an autobiography.
The expected sample proportion of customers who purchased an autobiography.
The standard deviation of the sample proportions of customers who purchased an autobiography.
Round your answer to three decimal places.
Three marbles are drawn with replacement from a bag containing six black and five grey marbles.
Let $X$X be the number of black marbles drawn.
Determine $E\left(X\right)$E(X).
Determine the standard deviation of $X$X.
Leave your answer in exact form.
Let $\hat{P}$^P be the proportion of black marbles drawn.
Determine $E\left(\hat{P}\right)$E(^P).
Determine the standard deviation of $\hat{P}$^P.
Leave your answer in exact form.