Answer:
There is a sufficient evidence to support the that machine 1 has the greater variance.
Step-by-step explanation:
The question with the complete details can be found online.
Start by stating the hypotheses.
The null hypothesis states that the variance of machine 1 is less than or equal to machine 2
So, we have:
[tex]H_o: \sigma_1 ^2 \le \sigma_2 ^2[/tex]
The alternate hypothesis will then be:
[tex]H_a: \sigma_1 ^2 > \sigma_2 ^2[/tex]
So, we have:
[tex]n_1 = 25[/tex] [tex]n_1 = 22[/tex]
Calculate the mean of Machine 1 and 2
The mean is:
[tex]\bar x =\frac{\sum x}{n}[/tex]
For machine 1, we have:
[tex]\bar x_1 =\frac{2.95+3.45+3.5+.....+3.12}{25}[/tex]
[tex]\bar x_1 =\frac{83.21}{25}[/tex]
[tex]\bar x_1 =3.3284[/tex]
For machine 2, we have:
[tex]\bar x_2 = \frac{3.22 + 3.3 + 3.34 + ..... +3.33}{22}[/tex]
[tex]\bar x_2 = \frac{72.12}{22}[/tex]
[tex]\bar x_2 = 3.2782[/tex]
Calculate the standard deviation of both machines
The standard deviation is:
[tex]\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}[/tex]
For machine 1, we have:
[tex]\sigma_1 = \sqrt{\frac{(2.95 - 3.3284)^2+(3.45 - 3.3284)^2+(3.5 - 3.3284)^2+.....+(3.12 - 3.3284)^2}{25-1}}[/tex]
[tex]\sigma_1 = \sqrt{\frac{(2.95 - 3.3284)^2+(3.45 - 3.3284)^2+(3.5 - 3.3284)^2+.....+(3.12 - 3.3284)^2}{24}}[/tex]
[tex]\sigma_1 = 0.2211[/tex]
For machine 2, we have:
[tex]\sigma_2 = \sqrt{\frac{(3.22 - 3.2782)^2+(3.3 - 3.2782)^2+(3.44 - 3.2782)^2+.....+(3.33 - 3.2782)^2}{22-1}}[/tex]
[tex]\sigma_2 = \sqrt{\frac{(3.22 - 3.2782)^2+(3.3 - 3.2782)^2+(3.44 - 3.2782)^2+.....+(3.33 - 3.2782)^2}{21}}[/tex]
[tex]\sigma_2 = 0.0768[/tex]
Calculate the degrees of freedom
[tex]df=n-1[/tex]
For machine 1;
[tex]df_1=25-1=24[/tex]
For machine 2
[tex]df_2=22-1=21[/tex]
Calculate the test statistic (t)
[tex]t = \frac{Var_1}{Var_2}[/tex]
Rewrite in terms of standard deviation
[tex]t = \frac{\sigma_1^2}{\sigma_2^2}[/tex]
[tex]t = \frac{0.2211^2}{0.0768^2}[/tex]
[tex]t = \frac{0.04888521}{0.00589824}[/tex]
[tex]t = 8.2881[/tex]
Lastly, calculate the p value.
This is the value of P(t > 8.2881) between the degrees of freedom i.e. 21 and 24
From the f distribution table
[tex]P(t > 8.2881) < 0.01[/tex]
Hence, we reject the null hypothesis
