Snow avalanches can be a real problem for travelers in the western United States
ID: 3150953 • Letter: S
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 = 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.)
(ii) Assume the slab thickness has an approximately normal distribution. Use a 10% 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.?
(b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution.
What is the value of the sample test statistic? (Round your answer to three decimal places.?
(c) Find the P-value. (Round your answer to four decimal places.?
Explanation / Answer
Answer (i)
Sample mean= x-bar = 61.81
Sample standard deviation = S = 10.65
Answer (ii)
a) Level of significance = alpha = 10% = 0.10
The null and alternative hypothesis is given as below:
Null hypothesis: H0: An average thickness for slab avalanches studied in a region of Canada is 66 cm.
Alternative hypothesis: Ha: An average thickness for slab avalanches studied in a region of Canada is different from 66 cm.
Symbolically
H0: µ = 66 versus Ha: µ 66
b) Here, we have to use the t distribution because we do not know the value for the population standard deviation and the sample size is less than 30.
The formula for t test statistic is given as below:
Test statistic = t = (sample mean – population mean) / [sample standard deviation / sqrt(n)]
Where n is the sample size.
We are given sample mean = 61.81
Sample standard deviation = 10.65
Population mean = 66
Sample size = n = 16
Now, plug all values in the above formula and calculate the value test statistic given as below:
Test statistic = t = (61.81 – 66) / [10.65 / sqrt (16) ]
Test statistic = t = -4.19/2.6625 = -1.57371
Test statistic = t = -1.574
c) The p-value is given as 0.1364 by using the t-table or t-distribution.
Summary:
t Test for Hypothesis of the Mean
Data
Null Hypothesis m=
66
Level of Significance
0.05
Sample Size
16
Sample Mean
61.81
Sample Standard Deviation
10.65
Intermediate Calculations
Standard Error of the Mean
2.6625
Degrees of Freedom
15
t Test Statistic
-1.5737
Two-Tail Test
Lower Critical Value
-2.1314
Upper Critical Value
2.1314
p-Value
0.1364
Do not reject the null hypothesis
t Test for Hypothesis of the Mean
Data
Null Hypothesis m=
66
Level of Significance
0.05
Sample Size
16
Sample Mean
61.81
Sample Standard Deviation
10.65
Intermediate Calculations
Standard Error of the Mean
2.6625
Degrees of Freedom
15
t Test Statistic
-1.5737
Two-Tail Test
Lower Critical Value
-2.1314
Upper Critical Value
2.1314
p-Value
0.1364
Do not reject the null hypothesis
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