Chapter 9.2, Problem 19P

Please provide the following information for Problems 11-22.

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

(b)Check Requirements What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. Compute the appropriate sampling distribution value of the sample test statistic.

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

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

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

Note: For degrees of freedom d.f. not given in the Student's t table, use the closest d.f. that is smaller. In some situations, this choice of d.f. may increase the P-value by a small amount and therefore produce a slightly more “conservative" answer.

Ski Patrol: Avalanches Snow avalanches can be a real problem for travelers in the western United States and Canada. A very common type of avalanche is called the slab avalanche. These have been studied extensively by David McClung, a professor of civil engineering at the University of British Columbia. Slab avalanches studied in Canada have an average thickness of $\mu =67$ cm (Source: Avalanche Handbook by D. McClung and P. Schaerer). The ski patrol at Vail, Colorado, is studying slab avalanches in its region.

A random sample of avalanches in spring gave the following thicknesses (in cm):

59	51	76	38	65	54	49	62
68	55	64	67	63	74	65	79

i. Use a calculator with mean and standard deviation keys to verify that $\bar{x}\approx \text{ 61}\text{.8 and }s\approx \text{ 10}\text{.6 cm}\text{.}$

ii. Assume the slab thickness has an approximately normal distribution. Use a 1% level of significance to test the claim that the mean slab thickness in the Vail region is different from that in Canada.

To determine

Whether the sample mean $\overline{x}=61.8$ and sample standard deviation $s=10.6$.

Answer to Problem 19P

Solution: Yes, the sample mean $\overline{x}=61.8$ and sample standard deviation $s=10.6$.

Explanation of Solution

To calculate the required statistics using the Minitab, follow the below instructions:

Step 1: Go to the Minitab software.

Step 2: Go to Stat > Basic statistics > Display Descriptive Statistics.

Step 3: Select ‘Thickness’ in variables.

Step 4: Click on OK.

The obtained statistics is:

Statistics

Variable	N	N*	Mean	SE Mean	StDev	Minimum	Q1	Median	Q3	Maximum
Thickness	16	0	61.81	2.66	10.65	38.00	54.25	63.50	67.75	79.00

From the Minitab output, the sample mean and sample standard deviation are approximately equals to $\overline{x}=61.8$ and $s=10.6$.

To determine

The level of significance, null and alternative hypothesis.

Answer to Problem 19P

Solution: The level of significance is $\alpha =0.01$. The null hypothesis is $H_0:\mu =67$ and alternative hypothesis $H_A:\mu \ne 67$.

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 $H_0:\mu =67$

Alternative hypothesis $H_A:\mu \ne 67$

To determine

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

Answer to Problem 19P

Solution: The student’s t distribution should be used. The sample test statistic is -1.96.

Explanation of Solution

Calculation:

We assume that x distribution is mound shape and symmetrical, because $\sigma$ is unknown, we can use student’s t distribution with $d.f=n-1=15$.

Using $\overline{x}=61.8,\mu =67,s=10.6,n=16$

The sample test statistic t is

$\begin{array}{l}t=\frac{(\overline{x}-\mu )}{\frac{s}{\sqrt{n}}}\\ t=\frac{(61.8-67)}{\frac{10.6}{\sqrt{16}}}\\ t=-1.9623\\ t\approx -1.96\end{array}$

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 19P

Solution: The P-value of the test statistic is 0.0688.

Explanation of Solution

Calculation:

We have t = -1.96

$P-\text{value}\text{=}P(t<-1.96)+P(t>1.96)$

Using Table 4 from the Appendix to find the specified area:

$0.05<P-\text{value}<0.10$

Graph:

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 ‘t’ and enter d.f = 15.

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

Step 5: Enter X-value as -1.96 and select ‘Both Tail’.

Step 6: Click on OK.

The obtained distribution graph is:

$\begin{array}{l}P-\text{value}\text{=}2(0.0344)\\ P-\text{value}\text{= 0}\text{.0688}\end{array}$

To determine

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

Answer to Problem 19P

Solution: The P-value $>\alpha$, hence we fail to reject the $H_0$. The data is not statistically significant for a level of significance of 0.01.

Explanation of Solution

The P-value of 0.0688 is greater than the level of significance ($\alpha$

) of 0.01. Therefore we don’t have enough evidence to reject the null hypothesis $H_0$, hence the data is not statistically significant for a level of significance of 0.01.

To determine

The interpretation for the conclusion.

Answer to Problem 19P

Solution: There is not enough evidence to conclude that the mean slab thickness in the Vail region is different from that in Canada.

Explanation of Solution

The P-value of 0.0688 is greater than the level of significance ($\alpha$

) of 0.01. Therefore we don’t have enough evidence to reject the null hypothesis $H_0$. There is not enough evidence to conclude that the mean slab thickness in the Vail region is different from that in Canada.