The variance in a production process is an important measureof the quality of the process. A large variance often signals anopportunity for improvement in the process by finding ways toreduce the process variance. Conduct a statistical test todetermine whether there is a significant difference between thevariances in the bag weights for the two machines. Use a .05 levelof significance. What is your conclusion? Which machine,if either,provides the greater opportunity for quality improvements?Machine1 2.95 3.45 3.50 3.75 3.48 3.26 3.33 3.20 3.16 3.20 3.22 3.38 3.90 3.36 3.25 3.28 3.20 3.22 2.98 3.45 3.70 3.34 3.18 3.35 3.12Machine2 3.22 3.30 3.34 3.28 3.29 3.25 3.30 3.27 3.38 3.34 3.35 3.19 3.35 3.05 3.36 3.28 3.30 3.28 3.30 3.20 3.16 3.33

Respuesta :

Answer:

[tex]F=\frac{s^2_1}{s^2_2}=\frac{0.2211^2}{0.0768^2}=8.28[/tex]

[tex]p_v =2*P(F_{24,21}>8.28)=7.22x10^{-6}[/tex]

Since the [tex]p_v < \alpha[/tex] we have enough evidence to reject the null hypothesis. And we can say that we have enough evidence to conclude that the variation between the two machines is significant at 5% of significance.  

Step-by-step explanation:

Data given and notation  

Machine 1: 2.95 3.45 3.50 3.75 3.48 3.26 3.33 3.20 3.16 3.20 3.22 3.38 3.90 3.36 3.25 3.28 3.20 3.22 2.98 3.45 3.70 3.34 3.18 3.35 3.12

Machine 2: 3.22 3.30 3.34 3.28 3.29 3.25 3.30 3.27 3.38 3.34 3.35 3.19 3.35 3.05 3.36 3.28 3.30 3.28 3.30 3.20 3.16 3.33

[tex]n_1 = 23 [/tex] represent the sampe size for machine 1

[tex]n_2 =22[/tex] represent the sample size for machine 2

[tex]\bar X_1 =3.33[/tex] represent the sample mean for machine 1

[tex]\bar X_2 =3.28[/tex] represent the sample mean for machine 2

[tex]s_1 = 0.2211[/tex] represent the sample deviation for machine 1

[tex]s^2_1 = 0.049[/tex] represent the sample variance for machine 1

[tex]s_2 = 0.0768[/tex] represent the sample deviation for machine 2

[tex]s^2_2 = 0.00690[/tex] represent the sample variance for machine 2

[tex]\alpha=0.05[/tex] represent the significance level provided

Confidence =0.95 or 95%

F test is a statistical test that uses a F Statistic to compare two population variances, with the sample deviations s1 and s2. The F statistic is always positive number since the variance it's always higher than 0. The statistic is given by:

[tex]F=\frac{s^2_1}{s^2_2}[/tex]

Solution to the problem  

System of hypothesis

We want to determine whether there is a significant difference between thevariances in the bag weights for the two machines , so the system of hypothesis are:

H0: [tex] \sigma^2_1 = \sigma^2_2[/tex]

H1: [tex] \sigma^2_1 \neq \sigma^2_2[/tex]

Calculate the statistic

Now we can calculate the statistic like this:

[tex]F=\frac{s^2_1}{s^2_2}=\frac{0.2211^2}{0.0768^2}=8.28[/tex]

Now we can calculate the p value but first we need to calculate the degrees of freedom for the statistic. For the numerator we have [tex]n_1 -1 =25-1=24[/tex] and for the denominator we have [tex]n_2 -1 =22-1=21[/tex] and the F statistic have 24 degrees of freedom for the numerator and 21 for the denominator. And the P value is given by:

P value

[tex]p_v =2*P(F_{24,21}>8.28)=7.22x10^{-6}[/tex]

And we can use the following excel code to find the p value:"=2*(1-F.DIST(8.2838,24,21,TRUE))"

Conclusion

Since the [tex]p_v < \alpha[/tex] we have enough evidence to reject the null hypothesis. And we can say that we have enough evidence to conclude that the variation between the two machines is significant at 5% of significance.