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 = 68 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	=	cm
s	=	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?


State the null and alternate hypotheses.

H₀: μ = 68; H₁: μ > 68 H₀: μ ≠ 68; H₁: μ = 68     H₀: μ = 68; H₁: μ < 68 H₀: μ < 68; H₁: μ = 68 H₀: μ = 68; H₁: μ ≠ 68

(b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution.

The Student's t, since we assume that x has a normal distribution and σ is unknown. The Student's t, since we assume that x has a normal distribution and σ is known.     The standard normal, since we assume that x has a normal distribution and σ is known. The standard normal, since we assume that x has a normal distribution and σ is unknown.

What is the value of the sample test statistic? (Round your answer to three decimal places.)


(c) Estimate the P-value.

P-value > 0.250 0.100 < P-value < 0.250     0.050 < P-value < 0.100 0.010 < P-value < 0.050 P-value < 0.010

Answer 1

i)

as we have given sample with sample size =n=16

then sample mean is given by

sample SD is given by

ii)

a)

Here we have to test at level of significance 1%

Hence level of significance is 1% (0.01)

we are interested in testing that if mean is different from 68 so

b)

here do not have knowledge about population standard deviation also sample size is small (less than 30) so we will use t distribution with df=n-1=16-1=15

so correct option is

The Student's t, since we assume that x has a normal distribution and σ is unknown.

test is given by

c)

since test is two tailed and t=-2.325 so P value is given by

P-Value=2P(t<-2.325)=2*0.0173=0.0346

so 0.010 < P-value < 0.050

since P value is more than 0.01 (level of significance) Hence we failed to reject H0 Hence there is no enough evidence to support the claim that mean is different from 68.