Question

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. Suppose slab avalanches studied in a region of Canada had an average thickness of μ = 66 cm. 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 sample mean and standard deviation keys to find x and s. (Round your answers to two decimal places.) x = 61.81 Correct: Your answer is correct. cm s = 10.64 Correct: Your answer is correct. cm (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 the region of Canada. (a) What is the level of significance? 0.100 Incorrect: Your answer is incorrect. State the null and alternate hypotheses. H0: μ = 66; H1: μ 66 H0: μ ≠ 66; H1: μ = 66 H0: μ < 66; H1: μ = 66 Incorrect: Your answer is incorrect.

Answer 1

Answer:

We conclude that the mean slab thickness in the Vail region is the same as that in the region of Canada.

Step-by-step explanation:

We are given that slab avalanches studied in a region of Canada had an average thickness of μ = 66 cm.

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

X: 59, 51, 76, 38, 65, 54, 49, 62, 68, 55, 64, 67, 63, 74, 65, 79.

Let $\mu$ = true mean slab thickness in the Vail region

So, Null Hypothesis, $H_0$ : $\mu$ = 66 cm      {means that the mean slab thickness in the Vail region is the same as that in the region of Canada}

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ 66 cm      {means that the mean slab thickness in the Vail region is different from that in the region of Canada}

The test statistics that will be used here is One-sample t-test statistics because we don't know about population standard deviation;

                             T.S.  =  $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$  ~ $t_n_-_1$

where, $\bar X$ = sample mean thickness = $\frac{\sum X}{n}$ = 61.81 cm

             s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 10.64

             n = sample of avalanches = 16

So, the test statistics =  $\frac{61.81-66}{\frac{10.64}{\sqrt{16} } }$  ~  $t_1_5$

                                    =  -1.575

The value of t-test statistics is -1.575.

Now, at a 1% level of significance, the t table gives a critical value of -2.947 and 2.947 at 15 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the mean slab thickness in the Vail region is the same as that in the region of Canada.