Question

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Answer 1

(a)

Objective:

To check whether the population mean times among the four machines are same or not at α=0.05 level of significance.

Null and alternative hypotheses:

The null and alternative hypotheses are as follows:

H₀: µ₁ = µ₂ = µ₃ = µ₄

That is, the mean time between breakdowns is same for the four machines.

H_a: µ₁ ≠ µ₂ ≠ µ₃ ≠ µ₄

That is, the mean time between breakdowns is not the same for the four machines.

Test statistic:

Procedure to obtain test statistic using Excel:

• Enter 4 machine data in 4 different columns.
• Go to Data > Data Analysis.
• Choose Anova: Single Factor.
• Click ok.
• In Input range, select the data in four columns.
• In Alpha, enter 0.05.
• Click ok.

Output is as follows:

Anova: Single Factor Count SUMMARY Groups Column 1 Column 2 Column 3 Column 4 Sum 44.4 55.2 59.4 68.4 Average 7.4 9.2 Variance 1.292 0.848 0.748 1.048 9.9 6 6 11.4 df ANOVA Source of Variation Between Groups Within Groups SS 49.605 19.68 MS 16.535 0.984 F P 16.8038618 -value F crit 0.00001 3.09839121 20 Total 69.285 23

From the output, the test statistic is 16.80 (with two decimal places) and p-value is 0.000 (with three decimal places).

Conclusion:

Since p-value is less than the significance level of 0.05, reject H₀ at 5% level of significance.

The correct option is third option. Reject H₀. There is sufficient evidence to conclude that the mean time between breakdowns is not the same for the four machines.

(b) Fisher’s LSD:

The mean of column 2 and column 4 is 9.2 and 11.4 respectively. The difference between these two means is calculated below:

Z-E = 92-11.4 =-22 E-|=22

From the output the degrees of freedom for within groups is 20 and MS for within groups is 0.984.

The null and alternative hypotheses are:

H₀: the mean time of machines 2 and 4 are same.

H_a: the mean time of machines 2 and 4 are not same.

Fisher’s LSD is calculated for machines 2 and 4 is as follows:

= 2.093/ + 166 From Percentage points of t Distribution table, 10.025.19 = 2.093 =1.1987

Since |x₂-bar – x₄-bar|> LSD_2,4, the null hypothesis rejected. Therefore, there is sufficient evidence to conclude that the mean time of machines 2 and 4 are not same.