Answer to Question #200775 in Statistics and Probability for husna

Question #200775

A human resources manager for a car company wanted to know whether production-line workers have more days absent than office workers. He took a random sample of eight workers from each category and recorded the number of days absent the previous year. Can we infer that there is a difference in days absent between the two groups of workers? (at 10% significance level) Production-line workers 4 0 6 8 3 11 13 5 Office workers 9 2 7 1 4 7 9 8


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Expert's answer
2021-06-01T17:35:45-0400

Null hypothesis: H0: μ1 - μ2 = Δ

Alternative hypothesis H1: μ1 - μ2 ≠ Δ

μPw=4+0+6+8+3+11+13+58=6.25\mu_{Pw}=\frac{4+0+6+8+3+11+13+5}{8}=6.25

μOw=9+2+7+1+4+7+9+88=5.875\mu_{Ow}=\frac{9+2+7+1+4+7+9+8}{8}=5.875

σPw=(46.375)2+(06.375)2+(66.375)2+...+(56.375)28=4.268\sigma_{Pw}=\sqrt{\frac{(4-6.375)^2+(0-6.375)^2+(6-6.375)^2+...+(5-6.375)^2}{8}}=4.268

σOw=(95.875)2+(25.875)2+(75.875)2+...+(85.875)28=3.137\sigma_{Ow}=\sqrt{\frac{(9-5.875)^2+(2-5.875)^2+(7-5.875)^2+...+(8-5.875)^2}{8}}=3.137

SEPw=4.26828=6.440SE_{Pw}=\frac{4.268^2}{\sqrt{8}}=6.440

SEOw=3.13728=3.479SE_{Ow}=\frac{3.137^2}{\sqrt{8}}=3.479

Apply t-distribution:

t=6.255.87504.26828+3.13728=0.200t=\dfrac{6.25-5.875-0}{\sqrt{\frac{4.268^2}{8}+\frac{3.137^2}{8}}}=0.200

We have DF = ((8-1)+(8-1))-1=13

Using a table below we can conclude that p-value for 1-tailed test is between 0.4 and 0.5, let's say 0.45. Then, for a 2-tailed test it's doubled, and p=0.9. It is more then 0.1, so there's no significant difference between the results.




(Using a p-score calculator we can see that p=0.84p=0.84, which is more than 0.1, so)

We can't say that there's a significant differense between the results.

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