Question #282297

. 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


1
Expert's answer
2022-02-21T15:23:58-0500

n1=n2=10s1=2s2=1.5α=0.05n_1=n_2=10\\s_1=2\\s_2=1.5\\\alpha=0.05

Hypothesis:

H0:σ1=σ2H_0:\sigma_1=\sigma_2 , there is no more variation in the number of rejects per day for the XX company

Ha:σ1>σ2H_a:\sigma_1>\sigma_2 , there is more variation in the number of rejects per day for the XX company


test statistic:

F=s12s22=221.52=1.7778F=\frac{s_1^2}{s_2^2}=\frac{2^2}{1.5^2}=1.7778

df1=n11=101=9df2=n21=101=9df_1=n_1-1=10-1=9\\df_2=n_2-1=10-1=9


critical value:

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


Reject H0H_0 if F>F0.05,9,9F\gt F_{0.05,9,9}


Since F=1.7778<F(0.05,9,9)=3.1789F=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.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS