Plz do this by hand with shown the equation but clear handwriting and plz don\'t
ID: 3356675 • Letter: P
Question
Plz do this by hand with shown the equation but clear handwriting and plz don't using excel is testing A particular component has four different failure modes. A manufacturer different design configurations (A·B, and C) that each utilize the component. A manufacturing engineer collects the following data on the number of failures that 2. in each mode for each configuration Mode 1 Mode 2 Mode 3 Mode 4 20 Configuration A 18 Configuration B 12 Configuration C6 The engineer wantsto determine whether the proportion of errorsis different 39 32 15 15 8 between the three configurations ate the appropriate null hypothesis and alternate hypothesis b) Compute the expected values under the null hypothesis c) Compute the value of the chi-square statistic s d) Find the p-value. What do you conclude? Note that if you cannot find the specific p- value, report the range from the table that justifies your response.Explanation / Answer
a) H0: Configuration A, B and C are independent.
H1: Configuration A, B and C are dependent of each other.
b)
Mode 1
Mode 2
Mode 3
Mode 4
Total
Configuration A
18
39
20
7
84 (R1)
Configuration B
12
32
15
8
67 (R2)
Configuration C
6
15
5
11
37 (R3)
Total
36 (C1)
86 (C2)
40 (C3)
26 (C4)
188 (n)
Expected values are found using :
ejk= RjCk/n
where Rj : Total for row j
Ck : total for column k
n = sample size
all three parameter are shown in above table .
Below table shows the expected values for each cell.
Mode 1
Mode 2
Mode 3
Mode 4
Total
Configuration A
16.08
38.42
17.87
11.62
84 (R1)
Configuration B
12.23
30.65
14.25
9.26
67 (R2)
Configuration C
7.08
16.92
7.87
5.12
37 (R3)
Total
36 (C1)
86 (C2)
40 (C3)
26 (C4)
188 (n)
like for config A mode 1 expected value is found using (67 x 36)/188 = 12.23.
c) Chi2 = ((oi - ei)2/ei)
where o : observed value
e : expected value
chi2 = (18-16.08)2/16.08 + (39-38.42)2/38.42 + .........................+ (11-5.12)2/5.12
= 10.785
d) for finding p value we need df value i.e, degree of freedom
df = (r-1)(c-1) r: number of rows ; c: number of columns
df = (3-1)(4-1)
= 6
we are assuming alpha = 0.05 or at 95% confidence interval.
Critical value or p- value at alpha = 0.05 and df= 6 from the chi square distribution table is 12.59
We know that we reject Null hypothesis if our calculated chi square value is greater than the p-value.
In part c , our chi square value was 10.785 which is less than the p- value of 12.59, therefore we will accept the null hypothesis and conclude that the three configurations are independent of each other.
Mode 1
Mode 2
Mode 3
Mode 4
Total
Configuration A
18
39
20
7
84 (R1)
Configuration B
12
32
15
8
67 (R2)
Configuration C
6
15
5
11
37 (R3)
Total
36 (C1)
86 (C2)
40 (C3)
26 (C4)
188 (n)