Question

Cathy Keene has just started a new business for design and manufacturing of reusable protective face masks. Due to the high demand, Cathy wants to invest in a new production machine. She considers Machine K and Machine S, but before making the final decisions, Cathy wants to compare their errors. To do so, Cathy asks the manufacturers to provide a sample of the recorded number of errors per day.

Machine K

5

3

8

7

4

3

3

5

6

4

Machine S

7

8

7

3

2

5

6

2

At 0.05 significance level, test whether there is a significant difference between the variances in the number of errors of the two machines.

Use the above information to answer the following questions:

The required test statistics is: Answer .

This a Answer –tail test.

The sample mean number of errors for Machine K is Answer

The sample mean number of errors for Machine S is Answer

The sample standard deviation for Machine K is Answer (rounded to four decimal places)

The sample standard deviation for Machine S is Answer (rounded to four decimal places)

The critical value is Answer

The calculated test statistics is Answer (rounded to four decimal places)

The statistical decision is: Answer

The conclusion is: Answer

Answer 1

For Machine K :

∑x = 48

∑x² = 258

n1 = 10

Mean , x̅1 = Ʃx/n = 48/10 = 4.8

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(258-(48)²/10)/(10-1)] = 1.7512

For Machine S :

∑x = 40

∑x² = 240

n2 = 8

Mean , x̅2 = Ʃx/n = 40/8 = 5

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(240-(40)²/8)/(8-1)] = 2.3905

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The required test statistics is: F tests for two variance

This a two tailed test.

The sample mean number of errors for Machine K is 4.8

The sample mean number of errors for Machine S is 5

The sample standard deviation for Machine K is 1.7512

The sample standard deviation for Machine S is 2.3905

The critical value:

Degree of freedom:  

df₁ = 8-1 = 7

df₂ = 10-1 = 9

Critical value(s):  

Lower tailed critical value, Fα/₂ = F.INV(0.05/2, 7, 9) = 0.2073

Upper tailed critical value, F₁-α/₂ = F.INV(1-0.05/2, 7, 9) = 4.1970

The calculated test statistics:

F = s₂²/s₁² = 2.3905² / 1.7512² = 1.8634

The statistical decision is: Fail to reject the null hypothesis.

The conclusion is:

There is not a significant difference between the variances in the number of errors of the two machines at 0.05 significance level.