- the mean trout lengths based on samples of size $5$.
- the average SAT score of a sample of high school students.
- the average male height based on samples of size $30$.
- the heights of college students at a sampled university
- all mean trout lengths in a sampled lake.
In this question, we need to choose the statements which best describe the sampling distribution.
A population refers to the whole group about which the conclusions are drawn. A sample is a particular group from which the data is collected. The sample size is always less than the population size.
A sampling distribution is a statistic that calculates the likelihood of an event based on data from a small subset of a larger population. It represents the frequency distribution of how far apart various outcomes will be for a particular population and is also called a finite-sample distribution. It relies upon several factors, including the statistic, sample size, sampling process, and overall population. It is used to calculate statistics for a given sample such as mean, range, variance, and standard deviation.
Inferential statistics require sampling distributions because they make it easier to understand a specific sample statistic regarding other possible values.
Expert Answer
In this question:
The mean trout lengths based on samples of size $5$,
The average male height based on samples of size $30$,
both are possible sampling distributions because they are samples drawn from a population.
However, in the statements,
Average SAT score of a sample of high school students,
Heights of college students at a sampled university,
All mean trout lengths in a sampled lake,
Average SAT score, heights of college students, and all mean trout lengths are approximated as population.
Hence, mean trout lengths based on samples of size $5$
and average male height based on samples of size $30$ are the correct examples of the sampling distribution.
The sampling distribution of sample proportions is discussed in the following examples to have a better understanding of the sampling distribution.
Example 1
Assume that $34\%$ of people own a smartphone. If a random sample of $30$ people is taken, find the probability that the proportion of samples who owned smartphones is between $40\%$ and $45\%$.
In this problem we have the following data:
Mean $=\mu_{\hat{p}}=p=0.34$
$n=30$.
Since, $np=(30)(0.34)=10.2$ and $n(1-p)=30(1-0.34)=19.8$ are greater than $5$, so we can say that $\hat{p}$ has the sampling distribution which is approximately normal with mean $\mu=0.34$ and standard deviation:
$\sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{30}}=\sqrt{\dfrac{0.34(1-0.34)}{30}}=0.09$
And so,
$P(0.4<\hat{p}<0.45)$ $=P\left(\dfrac{0.4-0.34}{0.09}<\dfrac{\hat{p}-p}{\sigma_{\hat{p}}}<\dfrac{0.45-0.34}{0.09}\right)$
$\approx P(0.67<Z<1.22)$
$=P(Z<1.22)-P(Z<0.67)$
$=0.3888-0.2486$
$=0.1402$
Example 2
Consider the data in example 1. If a random sample of $63$ people was surveyed, what is the probability that more than $40\%$ of them own a smartphone?
Since,
$np=63(0.34)=21.42$ and $n(1-p)=63(1-0.34)=41.58$ are greater than $5$, therefore the sampling distribution of sample proportion is approximately normal with mean $\mu=0.34$ and standard deviation:
$\sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{63}}=\sqrt{\dfrac{0.34(1-0.34)}{63}}=0.06$
So, $P(\hat{p}>0.4)=\left(\dfrac{\hat{p}-p}{\sigma_{\hat{p}}}>\dfrac{0.4-0.34}{0.06}\right)$
$\approx P(Z>1)$
$=1-P(Z<1)$
$=1-0.3413$
$=0.6587$