# Answer in Statistics and Probability for Jaja #251702

“n_1=10 \nnmean = bar{x_1} = frac{1667+16.29+…+15.88+16.99}{10} = 16.339 \nnsigma_1 = 0.020 \nnn_2 = 10 \nnbar{x_2} = frac{16.19+16.49+…+16.39+16.91}{10}=16.14 \nnsigma_2 = 0.025”

Now, if two machines are used for filling plastic bottles with a net volume of 16.0 ounces , then the net mean difference between the two machines would be zero

“H_0: mu_1 -mu_2 = 0 \nnH_1: mu_1 -mu_2 u2260 0”

It is a two-tailed test

α=0.10

Critical value

“Z_c = 1.645”

Reject H_{0} if |Z| > 1.645

Test-statistic:

“Z = frac{bar{x_1} – bar{x_2}}{sqrt{frac{sigma^2_1}{n_1} + frac{sigma_2^2}{n_2}}} \nnZ = frac{16.339-16.14}{sqrt{frac{0.02^2}{10} + frac{0.025^2}{10}}} \nnZ = -19.66”

Here we noticed that “|z| = 19.656 > Z_c = 1.645”

Hence, null hypothesis is rejected

Conclusion: There is enough evidence to claim that the two machines which are used for filling plastic bottles do not have a net volume of 16.0 ounces, at the “alpha = 0.1” significance level