Chapter 10.3, Problem 18E

To determine

a. To state:

The null and alternative hypothesis.

Answer to Problem 18E

Solution:

The null hypothesis is $H_0:\mu \ge 33,600$ and alternative hypothesis is $H_a:\mu <33,600$.

Explanation of Solution

Consider the following scenario,

“Teachers’ salaries in one state are very low, so low that educators in that state regularly complain about their compensation. The state mean is $33, 600, but teachers in one district claim that the mean in their district is significantly lower. They survey a simple random sample of 22 teachers in the district and calculate a mean salary of $32, 400 with a standard deviation of $1520”.

State the null and alternative hypothesis.

Suppose teachers claims that mean in their district is significantly lower than state mean of $33, 600. That is the research hypothesis mathematically as $\mu <33600$. The logical opposite is $\mu \ge 33,600$. Thus the hypotheses are tested as follows,

The Null hypothesis: $H_0:\mu \ge 33,600$

The alternative hypothesis: $H_a:\mu <33,600$

Final statement:

The null hypothesis is $H_0:\mu \ge 33,600$ and alternative hypothesis is $H_a:\mu <33,600$.

To determine

b. To determine:

Which distribution to use for the test statistic, and state the level of significance.

Answer to Problem 18E

Solution:

The t-test statistic is used and the level of significance of $\alpha =0.01$.

Explanation of Solution

The t-test statistic is appropriate to use in this case because the claim is about a population mean, the populations normally distributed, the population standard deviation is unknown, and the sample is a simple random sample. The level of significance is $\alpha =0.01$.

So, t-test statistic is used and the level of significance of $\alpha =0.01$.

Final statement:

The t-test statistic is used and the level of significance of $\alpha =0.01$.

To determine

c. To gather:

The data and calculate the necessary sample statistics.

Answer to Problem 18E

Solution:

The calculated test statistic is -3.7030.

Explanation of Solution

The test statistic formula:

Test statistic for a hypothesis test for a population mean ($\sigma$ Unknown):

When the population standard deviation is unknown, the sample taken is a simple random sample, and either the sample size is at least 30 or the population distribution is approximately normal, the test statistic for a hypothesis test for a population mean is given by,

$t=\frac{\overline{x}-\mu }{\left(\frac{s}{\sqrt{n}}\right)}$

Where n is the sample size,

$\overline{x}$ is the sample mean,

$\mu$ is the presumed value of the population mean from the null hypothesis, and

$s$ is the sample standard deviation.

The given information is,

$\overline{x}=\$32,400,s=\$1520,n=22\text{and}\mu =\$33,600$.

Substitute these values in the test statistic formula to get the following,

$\begin{array}{rll}t& =& \frac{\overline{x}-\mu }{\left(\frac{s}{\sqrt{n}}\right)}\\ & =& \frac{32400-33600}{\left(\frac{1520}{\sqrt{22}}\right)}\\ & =& \frac{-1200}{\left(\frac{1520}{4.6904}\right)}\\ & =& \frac{-1200}{324.0651}\\ & =& -3.7030\end{array}$

So, the calculated test statistic is -3.7030.

Final statement:

The calculated test statistic is -3.7030.

To determine

d. To draw:

To Draw a conclusion and interpret the decision.

Answer to Problem 18E

Solution:

There is a significance evidence to support the teachers’ claims that the mean in their district is significantly lower at the 0.01 level of significance.

Explanation of Solution

Assume that the population standard deviation is unknown and the population distribution is approximately normal.

Rejection Region for Hypothesis Tests for Population Means ($\sigma$ Unknown):

If p-value $\le \alpha$, then reject the null hypothesis.

If p- value $>\alpha$, then fail to reject the null hypothesis.

Degrees of Freedom for t in a Hypothesis Test for a Population Means ($\sigma$ Unknown):

In a hypothesis test for a population mean where the population standard deviation is unknown, the number of degrees of freedom for the Student’s t- distribution of the test statistic is given by

$df=n-1$

Where n is the sample size.

Form the given information this is the left-tailed test since $H_a:\mu <33,600$.

From the given information n = 22 then the degrees of freedom is given below,

$\begin{array}{rll}df& =& n-1\\ & =& 22-1\\ & =& 21\end{array}$

Conclusion:

Consider the level of significance of $\alpha =0.01$ and $t=-3.7030$.

Use the “p value of t-test statistic” calculator to get the following,

$p-\text{value}=0.0007$

The p -value is approximately 0.0007. Since $p-\text{value}\le \alpha \Rightarrow 0.0007\le 0.01$.

The diagrammatic representation is given below,

Thus the conclusion is to reject the null hypothesis.

So, there is significance evidence to support the teachers’ claims that the mean in their district is significantly lower at the 0.01 level of significance.

Final statement:

There is a significance evidence to support the teachers’ claims that the mean in their district is significantly lower at the 0.01 level of significance.