Question

A quality control manager suspects that the quality of items that are manufactured on a Monday is better than that of items manufactured on a Wednesday. In a random sample of 400 items manufactured on a Monday, 370 were rated acceptable or better, and in a random sample of 300 items manufactured on Wednesday, 260 were rated as acceptable or better. Can you conclude that the true proportion of items rated acceptable or better is greater on Monday than on Wednesday? Use a significance level of 5%. Assume the items produced on different days are independent. Use p-value method. The data are summarized in the following contingency table. (19 points total)

	Acceptable or Better	Not Acceptable
Monday	370	30
Wednesday	260	40

Label the parameters: (3 points)

State null and alternative hypotheses in parameters notation: (2 points)

H₀:

H_a:

Propose an appropriate hypothesis test: (2 points)

Verify the required conditions for the proposed hypothesis test: (4 points)

Independence and randomization assumption:

Normality assumption:

Report a test statistic and p-value: (4 points)

Test statistic =

p-value =

Note: Attach R codes and outputs in the following space:

           

         

Make a decision of hypothesis test: (2 points)

The significant p-value is less than 0.05 and so reject the null hypothesis at α = 0.05.

Make a conclusion of hypothesis test in context: (2 points)

Answer 1

p₁ : proportion of item rated acceptable on Monday

p₂ : proportion of item rated acceptable on Wednesday

H₀ : proportion of item rated acceptable on Monday is equal as item rated acceptable on wednesday

against

H₁ : proportion of item rated acceptable on Monday is greater than wednesday

H₀ : p₁=p₂. Vs. H₁ : p₁>p₂

Or

H₀ : p₁-p₂=0. Vs. H₁ : p₁-p₂>0

Assumptions

1) Both samples of item are manufactured on different days therefore both samples are Independent and random ..

2) sample sizes are 400 and 300 which is greater than 30 therefore normality assumption verified

Test Statistic :

  

n₁ =400 , n₂=300, p₁=370/400 = 0.925, p₂=260/300 = 0.867

p=(370+260)/(400+300) = 0.9, 1-p= 0.1,

1/n₁=1/400=0.0025, 1/n₂=1/300= 0.0033

Z=(0.925-0.867)/sqrt(0.9*0.1*(0.0025+0.0033))

Z= 0.058/0.02284

Z=2.5394

let calculate p-value = 0.005703

from above we can see that p value is less than 0.05

therefore we reject H₀

Conclusion : proportion of item rated acceptable on Monday is greater than on Wednesday i.e items manufactured on Monday is better than items manufactured on Wednesday.