Question

Teacher Salaries. Data on salaries in the public school system are published annually in Ranking of the States and Estimates of School Statistics by the National Education Association. The mean annual salary of (public) classroom teachers is $\backslash (55.4$thousand. Assume a standard deviation of $\backslash )9.2$thousand. Do the following tasks for the variable "annual salary" of classroom teachers.

a. Determine the sampling distribution of the sample mean for samples of size $64$Interpret your answer in terms of the distribution of all possible sample mean salaries for samples of $64$classroom teachers.

b. Repeat part (a) for samples of size$256$

c. Do you need to assume that classroom teacher salaries are normally distributed to answer parts (a) and (b)? Explain your answer.

d. What is the probability that the sampling error made in estimating the population means salary of all classroom teachers by the mean salary of a sample of $64$classroom teachers will be at most $\backslash (1000$?

e. Repeat part (d) for samples of size$\backslash )256$

Short answer

Part (a) Mean is$\$49.0$and the standard deviation is$\$1.15$

Part (b) Mean is $\$49.0$and the standard deviation is$\$0.575$

Part (c) Classroom teacher salaries are normally distributed to answer parts.

Part (d) The risk of making a sampling mistake in predicting the population mean wage of a sample of $64$classroom teachers $0.6156$is $\$1000$

Part (e) The probability of making a sampling error in estimating the population means a sample of $64$classroom instructors' income will be at most$0.9182$

Step-by-step solution

Part (a) Step 1: Given information

Teachers in (public) classrooms earn an average yearly income of $\$49.0$thousand dollars. Assume a $\$9.2$thousand standard deviation.

The "annual pay" of classroom teachers is the variable of concern here. We have $\mu =49$and $\sigma =9.2$based on the above data.

Part (a) Step 2: Concept

Formula used:

Part (a) Step 3: Calculation

For samples of size $64$, the sampling distribution of the sample mean is as follows: Size of the sample,$n=64$

Mean of $\overline{x}$

$$\begin{gathered} {\mu }_{\overline{x}}=\mu \\ =\$49.0 \end{gathered}$$

Standard deviation of $\overline{x}$

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}} \\ =\frac{9.2}{\sqrt{64}} \\ =\frac{9.2}{8} \\ =\$1.15 \end{gathered}$$

The mean and standard deviation of all possible sample mean wages for samples of $64$class teachers are $\$49$and $\$1.15$, respectively, for samples of size$64$

Part (b) Step 1: Calculation

For samples of size $256$, the sampling distribution of the sample mean is as follows: The sample size is$n=256$

Mean of $\overline{x},{\mu }_x=\mu$

$=\$49.0$

Standard deviation of $\overline{x}$,

$$\begin{gathered} {\sigma }_{\overline{x}}==\frac{\sigma }{\sqrt{n}} \\ =\frac{9.2}{\sqrt{256}} \\ =\frac{9.2}{16} \\ =\$0.575 \end{gathered}$$

The mean and standard deviation of all potential sample means salaries for samples of $64$class teachers are $\$49.0$thousand and $\$0.575$thousand, respectively, for samples of size$64$

Part (c) Step 1: Explanation

(c) There's no need to presume that classroom instructor salaries are spread evenly throughout answer parts.

(a) and (b) are due to the huge size of the samples$\left(n\ge 30\right)$

Part (d) Step 1: Calculation

The likelihood that the average classroom teacher wage will be at least $\$1000$,

$$\begin{gathered} P\left(\right|\overline{X}-\mu |\le 1)=P\left(\frac{|\overline{X}-\mu |}{\sigma /\sqrt{n}}\le \frac{1}{9.2/\sqrt{64}}\right) \\ =P\left(\left|Z\right|<\frac{1}{1.15}\right) \\ =P(-0.87<Z<0.87) \\ =P(Z<0.87)-P(Z<-0.87) \\ =0.8078-0.1922 \\ =0.6156 \end{gathered}$$

As a result, the risk of making a sampling mistake in predicting the population mean wage of a sample of $64$classroom teachers $\$1000$is$0.6156$

Part (e) Step 1: Calculation

Suppose samples of size,$n=256$

The likelihood that the average classroom teacher wage will be at least $\$1000$,

$$\begin{gathered} P\left(\right|\overline{X}-\mu |\le 1)=P\left(\frac{|\overline{X}-\mu |}{\sigma /\sqrt{n}}\le \frac{1}{9.2/\sqrt{256}}\right) \\ =P\left(\left|Z\right|<\frac{1}{0.575}\right) \\ \approx P\left(\right|Z|<1.74) \\ =P(-1.74<Z<1.74) \end{gathered}$$

$$\begin{gathered} =P(Z<1.74)-P(Z<-1.74) \\ =0.9591-0.0409 \\ =0.9182 \end{gathered}$$

As a result, the probability of making a sampling error in estimating the population means a sample of $256$classroom instructors' income $0.9182$will be at most $\$1000$