Which plan to prove △HML≅△HMK CANNOT be used based on the information in the diagram?
1 answer:
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Answer:
p value = 0.0174
Conclusion : we reject the null hypothesis
Step-by-step explanation:
Thinking step:
We need to perform a test to determine if the proportion of the good parts is the same for all three shifts at a significance level of
Assumption : all the population for each or the three shifts is not equal.
Calculation:
Let p₁ be the sample of the first shift
p₂ be the sample of the second shift
p₃ be the sample of the third shift
According to the null hypothesis
H₀ = p₁ = p₂ = p₃
In other words, all the population sample proportions are equal.
Alternatively, we can assume that the three shift are not equal p₁ ≠ p₂ ≠ p₃
Tabulating and performing the ² test gives 8.10
degrees of freedom:
df = k - 1
= 3 - 1 = 2
Thus the degree of freedom is 2
Solving using the MINITAB software gives: p = 0.174
The solution shows that the p value < level of significance, then p-value lies in the range 0.0174≤≤0.05
Therefore, we reject the null hypothesis based on the fact that the three shifts are not equal.