Answer to Question #282297 in Statistics and Probability for abeni

Question

. XX company assembles electrical components for the last 10 days the company had averaged 9 rejected with standard deviation of 2 rejects.DD company averaged 8.5 rejected with standard deviation of 1.5 rejects. At the 0.05 level of significance can we conclude that there is more variation in the number of rejects per day for the XX company

Answer 1

$n_1=n_2=10\\s_1=2\\s_2=1.5\\\alpha=0.05$

Hypothesis:

$H_0:\sigma_1=\sigma_2$ , there is no more variation in the number of rejects per day for the XX company

$H_a:\sigma_1>\sigma_2$ , there is more variation in the number of rejects per day for the XX company

test statistic:

$F=\frac{s_1^2}{s_2^2}=\frac{2^2}{1.5^2}=1.7778$

$df_1=n_1-1=10-1=9\\df_2=n_2-1=10-1=9$

critical value:

$F_{(0.05,9,9)}=3.1789$

Reject $H_0$ if $F\gt F_{0.05,9,9}$

Since $F=1.7778<F_{(0.05,9,9)}=3.1789$ we accept null hypothesis. There is no more variation in the number of rejects per day for the XX company at 5% significance level.