10.2 Problem 10.1 continued[the percentage of ash in coal from region 1 has popu
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Question
10.2 Problem 10.1 continued[the percentage of ash in coal from region 1 has population mean 15 and population standard deviation 4.30. The percentage of ash in coal from region 2 has a normal distribution with unknown mean, , and unknown standard deviation, . In a random sample of coal from region 2, n = 20, = 12.12 and s= 3.27]
(a) With = .05, test to see if the mean for region 2, i.e., , differs from 15, which is the mean for region 1. Define a parameter to be used in determining the hypotheses for this testing situation and state the null and alternative hypotheses to be tested in this situation.
(b) What conclusion would be reached? State your conclusion in the terminology of this example; not just reject some hypothesis or do not reject a hypothesis. Hint: For region 2, is unknown.
(c) For the test in parts (a) and (b), bound its p-value.
(d) How large of a sample is needed so that for the test in part (a), (13) = .01? Hint: Because is unknown for region 2, you may use its estimate and you may round to two-decimal places.
(e) Based on your answer to part (b), for the test in part(a), i.e., with n=20, is (13) < .01, (13) = .01, or (13) > .01?
Explanation / Answer
(a) With = .05, test to see if the mean for region 2, i.e., , differs from 15, which is the mean for region 1. Define a parameter to be used in determining the hypotheses for this testing situation and state the null and alternative hypotheses to be tested in this situation.
Solution:
Here, we have to use the one sample t test for the population mean. The null and alternative hypothesis for this test is given as below:
Null hypothesis: H0: The percentage of ash in coal from region 2 has a mean equal to 15.
Alternative hypothesis: Ha: The percentage of ash in coal from region 2 has a mean differs from 15.
H0: µ = 15 versus Ha: µ 15
We are given level of significance = alpha = = .05
µ = 15
Sample mean = xbar = 12.12
Sample standard deviation = S = 3.27
Sample size = n = 20
The test statistic formula is given as below:
Test statistic =t = (xbar – µ) / [S/sqrt(n)]
Test statistic = t = (12.12 – 15) / [3.27/sqrt(20)]
Test statistic = t = -3.9388
t Test for Hypothesis of the Mean
Data
Null Hypothesis m=
15
Level of Significance
0.05
Sample Size
20
Sample Mean
12.12
Sample Standard Deviation
3.27
Intermediate Calculations
Standard Error of the Mean
0.7312
Degrees of Freedom
19
t Test Statistic
-3.9388
Two-Tail Test
Lower Critical Value
-2.0930
Upper Critical Value
2.0930
p-Value
0.0009
Reject the null hypothesis
(b) What conclusion would be reached? State your conclusion in the terminology of this example; not just reject some hypothesis or do not reject a hypothesis. Hint: For region 2, is unknown.
Solution:
Here, we get the p-value as 0.0009 which is less than the given level of significance or alpha value 0.05, so we reject the null hypothesis that the percentage of ash in coal from region 2 has a mean equal to 15.
This means, we concluded that the percentage of ash in coal from region 2 has a mean differs from 15.
(c) For the test in parts (a) and (b), bound its p-value
Solution:
P-value = 0.0009
(d) How large of a sample is needed so that for the test in part (a), (13) = .01? Hint: Because is unknown for region 2, you may use its estimate and you may round to two-decimal places.
Sample Size Determination
Data
Population Standard Deviation
3.27
Sampling Error
1
Confidence Level
95%
Intemediate Calculations
Z Value
-1.9600
Calculated Sample Size
41.0763
Result
Sample Size Needed
42
(e) Based on your answer to part (b), for the test in part(a), i.e., with n=20, is (13) < .01, (13) = .01, or (13) > .01?
Solution:
(13) < .01
t Test for Hypothesis of the Mean
Data
Null Hypothesis m=
15
Level of Significance
0.05
Sample Size
20
Sample Mean
12.12
Sample Standard Deviation
3.27
Intermediate Calculations
Standard Error of the Mean
0.7312
Degrees of Freedom
19
t Test Statistic
-3.9388
Two-Tail Test
Lower Critical Value
-2.0930
Upper Critical Value
2.0930
p-Value
0.0009
Reject the null hypothesis
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