Chapter 9, Problem 8CRP

In Problems 8-12, please use the following steps (i)-(v) for all hypothesis tests:

(i) What is the level of significance? State the null and alternate hypotheses.

(ii) Check Requirements What sampling distribution will you use? What assumptions are you making? What is the value of the sample test statistic?

(iii) Find (or estimate) the P-value. Sketch the sampling distribution and show the area corresponding to the P-value.

(iv) Based on your answers in parts (i)-(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level a?

(v) Interpret your conclusion in the context of the application.

Note: For degrees of freedom d.f. not in the Student’s t table, use the closet d.f. that is smaller. In some situation, this choice of d.f. may increase the P-value a small amount and thereby produce a slightly more "conservative” answer.

Testing and Estimating µ with σ Known Let x be a random variable that represents micrograms of lead per liter of water (µg/L). An industrial plant discharges water into a creek. The Environmental Protection Agency has studied the discharged water and found x to have a normal distribution, with σ = 0.7 µg/L (Reference: EPA Wet lands Case Studies).

(a) The industrial plant says that the population mean value of x is µ = 2.0 µg/L. However. A random sample of n = 10 water samples showed that $\bar{x}=2.56\mu \text{g/L}$. Does this indicate that the lead-concentration population mean is higher than the industrial plant claims? Use α = 1%.

(b) Find a 95% confidence interval for µ using the sample data and the EPA value for σ.

(e) How large a sample should be taken to be 95% confident that the sample mean $\bar{x}$ is within a margin of error E = 0.2 µg/L of the population mean?

To determine

The level of significance, null and alternative hypothesis.

Answer to Problem 8CRP

Solution: The level of significance is $\alpha =0.01$. The null hypothesis is $H_0:\mu =2.0$ and alternative hypothesis $H_1:\mu >2.0$.

Explanation of Solution

The level of significance is defined as the probability of rejecting the null hypothesis when it is true, it is denoted by $\alpha =0.01$.

Null hypothesis $H_0:\mu =2.0$

Alternative hypothesis $H_1:\mu >2.0$

To determine

To find: The sampling distribution that should be used and compute the value of the sample test statistic.

Answer to Problem 8CRP

Solution: The normal distributionshould be used. The sample test statisticis 2.53.

Explanation of Solution

Calculation:

We will use the normal distribution because the original distribution x is normal and also $\sigma$ is known.

Using $\overline{x}=2.56,{\mu }_{\overline{x}}=2,\sigma =0.7,n=10$

The sample test statistic z is

$\begin{array}{l}z=\frac{\overline{x}-{\mu }_{\overline{x}}}{\frac{\sigma }{\sqrt{n}}}\\ z=\frac{2.56-2}{\frac{0.7}{\sqrt{10}}}\\ z=2.5298\\ z\approx 2.53\end{array}$

To determine

To find: The P-value of the test statistic and sketch the sampling distribution showing the area corresponding to the P-value.

Answer to Problem 8CRP

Solution: The P-value of the test statistic is 0.0057.

Explanation of Solution

Calculation:

We have z = 2.53

$\begin{array}{l}P-value=P(z>2.53)\\ =1-P(z\le 2.53)\\ \text{Using Table 3 from the Appendix to find the specified area:}\\ =1-0.9943\\ =0.0057\end{array}$

Graph:

To draw the required graphs using the Minitab, follow the below instructions:

To draw the required graphs using the Minitab, follow the below instructions:

Step 1: Go to the Minitab software.

Step 2: Go to Graph > Probability distribution plot > View probability.

Step 3: Select ‘Normal’ and Mean = 0, Standard deviation = 1.

Step 4: Click on the Shaded area > X value.

Step 5: Enter X-value as 2.53 and select ‘Right Tail’.

Step 6: Click on OK.

The obtained distribution graph is:

$P-\text{value}\text{=}0.0057$

To determine

Whether we reject or fail to reject the null hypothesisand whether the data is statistically significant for a level of significance of 0.01.

Answer to Problem 8CRP

Solution: The P-value < α, hence we will reject the $H_0$. The data isstatistically significant for a level of significance of 0.01.

Explanation of Solution

The P-value of 0.0057 is less than the level of significance (α) of 0.01. Therefore we have to reject the null hypothesis $H_0$, hence the data is statistically significant for a level of significance of 0.01.

To determine

The interpretation for the conclusion.

Answer to Problem 8CRP

Solution: There is enough evidence to conclude that lead concentration population average is higher than the industrial plant claims.

Explanation of Solution

The P-value of 0.0057 is less than the level of significance ($\alpha$) of 0.01. Therefore we have to reject the null hypothesis $H_0$. There is enough evidence to conclude that lead concentration population average is higher than the industrial plant claims

To determine

To find: A 95% confidence interval for the mean using the sample data.

Answer to Problem 8CRP

Solution:

The 95% confidence interval for $\mu$ is (2.13, 2.99).

Explanation of Solution

Calculation:

We have to find 95% confidence interval

$\begin{array}{l}\sigma =0.7\\ n=10\\ \text{Using}\text{Table}3\text{of}\text{Appendix}\\ z_c=1.96\\ E=z_c\frac{\sigma }{\sqrt{n}}\\ E=1.96\frac{0.7}{\sqrt{10}}\\ E=0.434\end{array}$

95% confidence interval is

$\begin{array}{l}\overline{x}-E<\mu <\overline{x}+E\\ 2.56-0.434<\mu <2.56+0.434\\ 2.13<\mu <2.99\end{array}$

The 95% confidence interval for $\mu$ is (2.13, 2.99).

To determine

To find: The minimum sample size so that the margin of error is not more than 0.2 $\mu g/L$.

Answer to Problem 8CRP

Solution:

The sample size is 48 so that the margin of error is not more than 0.2 $\mu g/L$.

Explanation of Solution

Calculation:

We have the sample size formula as follows

$\begin{array}{l}n={\left(\frac{z_c\sigma }{E}\right)}^2\\ \sigma =0.7\\ E=0.2\\ \text{Using}\text{Table}3\text{of}\text{Appendix}\\ z_c=1.96\end{array}$

$\begin{array}{l}n={\left(\frac{z_c\sigma }{E}\right)}^2\\ n={\left[\frac{(1.96)(0.7)}{0.2}\right]}^2\\ n=47.05\\ n\approx 48\end{array}$

Therefore n = 48 is the minimum sample size to ensure that the margin of error E is not more than 0.2 $\mu g/L$.