The following two tables are of inclusion sizes, particles embedded in a weld, f
ID: 3201023 • Letter: T
Question
The following two tables are of inclusion sizes, particles embedded in a weld, from two different shielding gasses. Perform these hypothesis tests (show calculations in a proper way):
Compare the two means.
Compare the two variances.
Follow these steps for each test:
Write the hypotheses clearly. In the hypotheses, use a mathematical notations. For example, do not say, “The means are the same.” Instead, write m1¹ m2 and give explanations of the variables.
Write the test statistic formula along with the formula of degree of freedom.
Calculate the Test Value.
Find the p-value.
Make a decision at the significance level of 5%.
Argon shielding
0.55
0.01
0.46
0.30
0.32
0.00
0.21
0.37
0.28
0.25
0.04
0.18
0.44
0.21
0.37
0.37
0.48
0.56
0.52
0.72
0.61
1.08
0.11
0.42
0.09
0.65
0.52
0.56
0.04
0.70
0.61
0.34
0.27
0.00
0.28
0.41
0.83
0.22
0.23
0.40
0.33
0.40
0.01
0.37
0.49
0.52
0.51
0.49
0.76
-0.20
Carbon dioxide shielding
0.34
0.41
0.23
0.30
0.52
0.37
0.41
0.49
0.59
0.11
0.25
0.36
0.78
0.06
0.63
0.41
0.21
0.39
0.70
0.55
0.23
0.54
0.66
0.48
0.33
0.34
0.11
0.00
0.03
0.47
0.85
0..30
0.28
0.45
0.45
0.37
0.14
0.61
0.79
0.43
0.47
0.73
0.39
0.63
0.53
0.85
0.46
0.22
0.14
0.55
Argon shielding
0.55
0.01
0.46
0.30
0.32
0.00
0.21
0.37
0.28
0.25
0.04
0.18
0.44
0.21
0.37
0.37
0.48
0.56
0.52
0.72
0.61
1.08
0.11
0.42
0.09
0.65
0.52
0.56
0.04
0.70
0.61
0.34
0.27
0.00
0.28
0.41
0.83
0.22
0.23
0.40
0.33
0.40
0.01
0.37
0.49
0.52
0.51
0.49
0.76
-0.20
Carbon dioxide shielding
0.34
0.41
0.23
0.30
0.52
0.37
0.41
0.49
0.59
0.11
0.25
0.36
0.78
0.06
0.63
0.41
0.21
0.39
0.70
0.55
0.23
0.54
0.66
0.48
0.33
0.34
0.11
0.00
0.03
0.47
0.85
0..30
0.28
0.45
0.45
0.37
0.14
0.61
0.79
0.43
0.47
0.73
0.39
0.63
0.53
0.85
0.46
0.22
0.14
0.55
Explanation / Answer
Solution
Let X represent the 'inclusion size' of Argon shielding and Y represent the 'inclusion size' of Carbon dioxide shielding.xi and yi represent the ith observation on X and Y respectively, i = 1,2, ....., n(50 in our case)
We assume: X N(µ1, 12) and Y N(µ2, 22)
Level of significance: 5% (given)
Test 1 for equality of variances
Null Hypothesis: H0: 12 = 22 vs alternative HA: 12 > 22
Test Statistics
F = s12/s22 where s12 and s22 are respetively the sample variances. (Note: the divisor in each case is 49 and not 50). Under the null hypothesis, F has an F-Distribution with degrees of freedom (49, 49).
Calculations
Mean X = 0.3738 Mean y = 0.4148 s12 = 0.0601 and s22 = 0.0471 F = 1.2749
Upper 5% critical value of F-Distribution with degrees of freedom (49, 49) ~ 1.40 [interpolated from Standard F-Table]
Decision
Since the calculated value of F is less than the table value, H0 is accepted. => the two variances are equal
i.e., 12 = 22 = 2, say
Test 2 for equality of means
Null Hypothesis: H0: µ1= µ2 vs alternative HA: µ1 µ2
Test Statistics
t = {(Mean X - Mean Y)}/{s(2/n)}where s = sq.rt{(s12 + s22)/2} Under the null hypothesis, t has an t -Distribution with degreesof freedom = (49 + 49) = 98.
Calculations
Mean X = 0.3738 Mean y = 0.4148 s = 0.2342 t = 1.7318
Upper 5% critical value of t-Distribution with degreesof freedom 98 = 1.98 [interpolated from Standard t-Table]
Decision
Since the calculated value of t is less than the table value, H0 is accepted. => the two means are equal
i.e., µ1= µ2.
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