Question

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 - 10,860; SSTR - 4,590 .. Set up the ANOVA table for this problem (to 2 decimals. If necessary). Round p-value to four decimal places Source of Variation Sum of Squares Degrees of Freedom Mean Square D-value Treatments 4590 0.16 0.9459 Error 6270 3135 x Total 10860 b. Use - .05 to test for any significant difference in the means for the three assembly methods. The p-value is between 0 and 10 What is your conclusion? Conclude

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Answer 1

Slep-1:- The sum et sauates for error is obtained as shown below: Given SS7 = 10,860 SSTR = 4590 SSE - SST - SSTR = 10,860 -4590 =6270 Step-d: from intormation given, there are three dibberent methods 457 assembling a poradeck, that iö 8=3 * 30 empbyees are randomly selected.. ni 30 . degree of freedom for treatments is, 2-1 of sSTR =8-1 -3-1 = 2 The degree of freedom de SSE = nor = 30°3 =27 for erroris

The degree of freedom for total is dt = n- .-30-1 Step-33 The mean sum of square for treatment is · MSTR = SSTR df SS TR 4590 = 2,295 The mean sum of senare for errol is. MSE - SSE disse 6270 97 =939.92 Step-4: The F Jalio iz obtained as shown below. FIR = MSTR MSE 2295 932.92 FTR = 9.882

The ANOVA Table is source to Variation sumeb Squades E de Mean Sea Msware Treatments 12. 9.882 14590 6270 · 127 2,295 232.22 ' Edി 10,860 29 Total Step-5:- State the hypotheses. Let the denote the mean for first method. Ma dencte the mean for and method. luz denote the mean for 3rd method. Null Hypothesis : Ho ill, = M2 = Uz Alternative Hypothesis; Hail, Ellz #3 Step-bi- The value of test Static is 9.89-2 because Cause Test static = FIR

Step. 7: The p-value obtained as shown below. The test statistic value is 9.882, the numerato degrees et boredom is a and the denominator degrees freedom is a7. gri 'F' distribution table' locate the column with' a' Under 'numerator degrees to Sereedom'. Locate the crow with number '27' under 'denominator degrees of Greco Locate the value that is close to the test statistic value. The value closer to test statistic (= 9.882) value is 9.08. The p-value Corresponding to the value 9:02 is 0.001. 1 P-value (= 0.001) is less than oool Step-8:- • The conclusion is stated below use the levelet significance 0-0.05 The p-value is less than 0.05 Here, the p-value is less than level of significance. That is, p-value (= less than 0.0l) <a (20.05). By the rejection sulle , the null hypothesy rejected. It can be concluded that, there is evidence that the mean for three methods is significantly different.