Question

The variance in a production process is an important measure of the quality of the process. A large variance often signals an opportunity for improvement in the process by finding ways to reduce the process variance. Conduct a statistical test to determine whether there is a significant difference between the variances in the bag weights for the two machines. Use a level of significance. What is your conclusion? Which machine, if either, provides the greater opportunity for quality improvements? Click on the datafile logo to reference the data. (to 4 decimals) (to 4 decimals) (to 2 decimals) - Select your answer -

Answer 1

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:

$H_o: \sigma_1 ^2 \le \sigma_2 ^2$

The alternate hypothesis will then be:

$H_a: \sigma_1 ^2 > \sigma_2 ^2$

So, we have:

$n_1 = 25$               $n_1 = 22$

Calculate the mean of Machine 1 and 2

The mean is:

$\bar x =\frac{\sum x}{n}$

For machine 1, we have:

$\bar x_1 =\frac{2.95+3.45+3.5+.....+3.12}{25}$

$\bar x_1 =\frac{83.21}{25}$

$\bar x_1 =3.3284$

For machine 2, we have:

$\bar x_2 = \frac{3.22 + 3.3 + 3.34 + ..... +3.33}{22}$

$\bar x_2 = \frac{72.12}{22}$

$\bar x_2 = 3.2782$

Calculate the standard deviation of both machines

The standard deviation is:

$\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}$

For machine 1, we have:

$\sigma_1 = \sqrt{\frac{(2.95 - 3.3284)^2+(3.45 - 3.3284)^2+(3.5 - 3.3284)^2+.....+(3.12 - 3.3284)^2}{25-1}}$

$\sigma_1 = \sqrt{\frac{(2.95 - 3.3284)^2+(3.45 - 3.3284)^2+(3.5 - 3.3284)^2+.....+(3.12 - 3.3284)^2}{24}}$

$\sigma_1 = 0.2211$

For machine 2, we have:

$\sigma_2 = \sqrt{\frac{(3.22 - 3.2782)^2+(3.3 - 3.2782)^2+(3.44 - 3.2782)^2+.....+(3.33 - 3.2782)^2}{22-1}}$

$\sigma_2 = \sqrt{\frac{(3.22 - 3.2782)^2+(3.3 - 3.2782)^2+(3.44 - 3.2782)^2+.....+(3.33 - 3.2782)^2}{21}}$

$\sigma_2 = 0.0768$

Calculate the degrees of freedom

$df=n-1$

For machine 1;

$df_1=25-1=24$

For machine 2

$df_2=22-1=21$

Calculate the test statistic (t)

$t = \frac{Var_1}{Var_2}$

Rewrite in terms of standard deviation

$t = \frac{\sigma_1^2}{\sigma_2^2}$

$t = \frac{0.2211^2}{0.0768^2}$

$t = \frac{0.04888521}{0.00589824}$

$t = 8.2881$

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

$P(t > 8.2881) < 0.01$

Hence, we reject the null hypothesis