Q. 7.1 - Elementary Statistics

Question

Population data: $2,3,5,5,7,8$

Part (a): Find the mean, $\mu$, of the variable.

Part (b): For each of the possible sample sizes, construct a table similar to Table $7.2$on the page $293$ and draw a dotplot for the sampling for the sampling distribution of the sample mean similar to Fig $7.1$on page $293$.

Part (c): Construct a graph similar to Fig $7.3$ and interpret your results.

Part (d): For each of the possible sample sizes, find the probability that the sample mean will equal the population mean.

Part (e):  For each of the possible sample sizes, find the probability that the sampling error made in estimating the population mean by the sample mean will be $0.5$ or less, that is, that the absolute value of the difference between the sample mean and the population mean is at most $0.5$.

Step-by-Step Solution

Answer

Part (a): The mean $\mu$ is $5$.

Part (b): When $n=1$,

When $n=2$,

When $n=3$,

When $n=4$,

When $n=5$,

When $n=6$,

Part (c): The dot plot is given below,

Part (d): The probability that the sample mean will equal the population mean are $\frac{1}{3},\frac{1}{5},\frac{1}{5},\frac{1}{5},\frac{1}{3},1$.

Part (e): The probability that the sampling error made in estimating the population are $\frac{1}{3},\frac{1}{3},\frac{2}{5},\frac{3}{5},\frac{2}{3},1$.

Part (a) Step 1. Given information

Consider the given question,

The population data is $2,3,5,5,7,8$.

Part (a) Step 2. Find the mean of the variable.

The mean $\mu$ is given below,

$$\begin{gathered} \mu =\frac{\sum _{x_i}}{N} \\ =\frac{2+3+5+5+7+8}{6} \\ =\frac{30}{6} \\ =5 \end{gathered}$$

Part (b) Step 1. Construct a table for n = 1 , 2 , 3 .

For each of the possible sample sizes, we construct a table.

If the sample size taken $n=1$,

If the sample size taken $n=2$,

If the sample size taken $n=3$,

Part (b) Step 2. Construct a table for n = 4 , 5 , 6 .

If the sample size taken $n=4$,

If the sample size taken $n=5$,

If the sample size taken $n=6$,

Part (c) Step 1. Construct the dot plot.

We will construct the dot plot for the sampling distribution of the sample mean.

To construct dot plot for the sampling distribution of the sample mean,

Part (d) Step 1. Find the probability that the sample mean will equal the population mean.

We can observe that from the dot plot there is two dot corresponding to $\mu =5$ when n is $1$.

Hence, the probability that sample mean will be equal to population mean$$\begin{gathered} =\frac{2}{6} \\ =\frac{1}{3} \end{gathered}$$

Similarly, the probability that sample mean will be equal to population mean when n is $2$ is $$\begin{gathered} =\frac{3}{15} \\ =\frac{1}{5} \end{gathered}$$ (As there are data-custom-editor="chemistry" $3$ dots corresponding $\mu =5$)

The probability that sample mean will be equal to population mean when n is $3$ is $$\begin{gathered} =\frac{4}{20} \\ =\frac{1}{5} \end{gathered}$$(As there are no dots corresponding $\mu =5$)

We can observe that from the dot plot there is one dot corresponding to $\mu =5$ when n is data-custom-editor="chemistry" $4$.

The probability that sample mean will be equal to population mean when n is $4$ is $$\begin{gathered} =\frac{3}{15} \\ =\frac{1}{5} \end{gathered}$$ (As there are $3$ dots corresponding $\mu =5$)

The probability that sample mean will be equal to population mean when n is $5$ is $\frac{2}{6}=\frac{1}{3}$.

The probability that sample mean will be equal to population mean for $n=6$ is $1$.

Part (e) Step 1. Find the probability that sampling error made in estimating the population mean.

Number of dots within $0.5$ or less of $\mu =5$ is $2$ out of $6$ when n is $1$.

Hence, the probability that $\overline{x}$ will be within $0.5$ or less of $\mu$ is $\frac{2}{6}=\frac{1}{3}$.

Number of dots within $0.5$ or less of $\mu$ is $5$ out of $15$ when n is $2$.

Hence, the probability that $\overline{x}$ will be within $0.5$ or less of $\mu$ is $\frac{5}{15}=\frac{1}{3}$.

Number of dots within $0.5$ or less of $\mu$ is $8$ out of $20$ when n is $3$.

Hence, the probability that $\overline{x}$ will be within $0.5$ or less of $\mu$ is $\frac{8}{20}=\frac{2}{5}$,

Number of dots within $0.5$ or less of $\mu$ is $9$ out of $15$ when n is $4$.

Hence, the probability that $\overline{x}$ will be within $0.5$ or less of $\mu$ is $\frac{9}{15}=\frac{3}{5}$.

Number of dots within $0.5$ or less of $\mu$ is $4$ out of $6$ when n is $5$.

Hence, the probability that $\overline{x}$ will be within $0.5$ or less of $\mu$ is $\frac{4}{6}=\frac{2}{3}$.

Number of dots within data-custom-editor="chemistry" $0.5$ or less of $\mu$ is $1$ out of $1$ when n is $6$.

Hence, the probability that $\overline{x}$ will be within $0.5$ or less of $\mu$ is $\frac{1}{1}=1$.