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. 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
Use {Exercise 13.07} Three different methods for assem = .05 to test for any significant difference in the means for the three assembly methods.
Calculate the value of the test statistic (to 2 decimals).
The p-value is Selectless than .01between .01 and .025between .025 and .05between .05 and .10greater than .10Item 11
What is your conclusion?
SelectConclude not all means of the three assembly methods are equalCannot reject the assumption that the means of all three assembly methods are equal
Item 12are my calculations correct?

Given :

Number of methods, k = 3

Number of the observations, n = 30

Degrees of freedom Treatment = k - 1

                                                     = 3 - 1 = 2

Degrees of freedom for error = n - k

                                                 = 30 - 3 = 27

       = 10800 - 4560

       = 6240

Source variation   Sum of square  Degrees of freedom  Mean square     F

Treatment                   4560                          2                         2280       9.8653

Error                            6240                         27                   231.11111111

Total                            10800

The critical value of F for the 0.05 sig level and df (227) is 3.354

Since the test stat > critical value of F, so we reject null hypothesis and state that there is a significant difference in the means of the three assembly methods.

l.ln.lln.k.nkl

step-by-step explanation:

amount of oatypop that contains the prize = 40%

amount   of oatypop which doesn't contain the prize = (100-40)=60%

probability that hannah will win a prize

    = s +f s +f f s+ff   f s+ff f f s+ infinity.where f= failure,s= success

=   or 60%

probability that hannah only has to buy 3 or less boxes before getting a prize= (s) and ( f s) and (ff s) and (f f f s)

        = = 87.04 %  

step-by-step explanation: