Three different methods for assembling a product were proposed by an industrial engineer. To investigate the number of units assembled correctly with each method, 30 employees were randomly selected and randomly assigned to the three proposed methods in such a way that each method was used by 10 workers. The number of units assembled correctly was recorded, and the analysis of variance procedure was applied to the resulting data set. The following results were obtained: SST = 10,800; SSTR = 4560.
1. Set up the ANOVA table for this problem (to 2 decimals, if necessary).
Source of Variation Sum of Squares Degrees of Freedom Mean Square F
Treatments
Error
Total
2. Use a= .05 to test for any significant difference in the means for the three assembly methods.
3. Calculate the value of the test statistic (to 2 decimals).
The p-value is:_______.
a. less than .01
b. between .01 and .025
c. between .025 and .05
d. between .05 and .10
e. greater than .10
4. What is your conclusion?
a. Not all means of the three assembly methods are equal
b. Cannot reject the assumption that the means of all three assembly methods are equal

Respuesta :

Answer:

is an attachment

test statistic = 9.87

p value is less than 0.01

Not all means of the three assembly methods are equal

Step-by-step explanation:

the anova table is an attachment

total number of methods = k = 3

number of observations n = 30

for treatment, df = 3-1 = 2

for error, df = 30 -3 (n-k) = 27

sst = 10800, sstr = 4560, sse = 10800-4560 = 6240

we find the mean of squares for error

sse/df of error = 6240/27 = 231.11

mean of squares for treatment = 4560/2 = 2280

test stat

F = 2280/231.11

= 9.8654

using f distribution table;

alpha = 0.05

df = 2 , 27

we get 3.35

h0; no difference in mean

h1; there is difference

9.87 > 3.35 so we reject H0, there is difference in means. all are not equal.

pvalue calculation

using excel,

FDIST(9.8654, 2, 27) = 0.0006078

p value is less than 0.01

we conclude that a. Not all means of the three assembly methods are equal

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