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djverab [1.8K]
3 years ago
12

Three different methods for assembling a product were proposed by an industrial engineer. To investigate the number of units ass

embled 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
Mathematics
1 answer:
In-s [12.5K]3 years ago
3 0

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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Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation : 

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<span> 3.1 </span>     Factoring: <span> a2 - 16</span> 

Theory : A difference of two perfect squares, <span> A2 - B2  </span>can be factored into <span> (A+B) • (A-B)

</span>Proof :<span>  (A+B) • (A-B) =
         A2 - AB + BA - B2 =
         A2 <span>- AB + AB </span>- B2 = 
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</span>Note : <span> <span>AB = BA </span></span>is the commutative property of multiplication. 

Note : <span> <span>- AB + AB </span></span>equals zero and is therefore eliminated from the expression.

Check : 16 is the square of 4
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Factorization is :       (a + 4)  •  (a - 4) 

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<span> 4.1 </span>   Find the Least Common Multiple 

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      The right denominator is :      <span> a-4 </span>

<span><span>                  Number of times each Algebraic Factor
            appears in the factorization of:</span><span><span><span>    Algebraic    
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    Denote the Right Multiplier by  R_Deno 

   Left_M = L.C.M / L_Deno = 1

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Making Equivalent Fractions :

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Add the two equivalent fractions which now have a common denominator

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      Least Common Multiple: 
      (a+4) • (a-4) 

Calculating Multipliers :

<span> 5.2 </span>   Calculate multipliers for the two fractions 


    Denote the Least Common Multiple by  L.C.M 
    Denote the Left Multiplier by  Left_M 
    Denote the Right Multiplier by  Right_M 
    Denote the Left Deniminator by  L_Deno 
    Denote the Right Multiplier by  R_Deno 

   Left_M = L.C.M / L_Deno = 1

   Right_M = L.C.M / R_Deno = a-4

Making Equivalent Fractions :

<span> 5.3 </span>     Rewrite the two fractions into<span> equivalent fractions</span>

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<span> 5.4 </span>      Adding up the two equivalent fractions 

(3a+8) - (2 • (a-4)) a + 16 ———————————————————— = ————————————————— (a+4) • (a-4) (a + 4) • (a - 4) <span>Equation at the end of step  5  :</span> a + 16 ————————————————— = 0 (a + 4) • (a - 4) <span>Step  6  :</span>When a fraction equals zero :<span><span> 6.1 </span>   When a fraction equals zero ...</span>

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the <span>denominator, </span>Tiger multiplys both sides of the equation by the denominator.

Here's how:

a+16 ——————————— • (a+4)•(a-4) = 0 • (a+4)•(a-4) (a+4)•(a-4)

Now, on the left hand side, the <span> (a+4) •</span> (a-4)  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
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Solving a Single Variable Equation :

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One solution was found :

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4 0
3 years ago
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