Someone help plZ and tY
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Answer:
Step-by-step explanation:
Hello!
You have two variables of interest
X₁: failure stress of a NiCrAlZr coating after nine 1-hr cycles.
X₂: failure stress of a NiCrAlZr coating after six 1-hr cycles.
a)
To be able to estimate the difference between the means using a confidence interval, you need that both variables have a normal distribution and to determine whether or not the population variances are equal.
If the population variances are equal, σ₁²=σ₂², you can use a pooled variance t-test
If the population variances are different, σ₁²≠σ₂², you have to use Welch's t-test
Using α: 0.05
The normality test for X₁ shows a p-value of 0.7449 ⇒ You can assume it has a normal distribution.
The normality test for X₂ shows a p-value of 0.9980 ⇒ You can assume it has a normal distribution.
The F-test for variance homogeneity shows a p-value of 0.6968 (H₀:σ₁²=σ₂²) ⇒You can assume both population variances are equal.
b) and c)
You need to test if both population means are the same, the hypotheses are:
H₀: μ₁=μ₂
H₁: μ₁≠μ₂
α: 0.05
The distribution of this test is a t with 13 degrees of freedom and the test is two-tailed, so to calculate the p-value you have to do the following:
P(t₁₃≤-4.23)+P(t₁₃≥4.23)= P(t₁₃≤-4.23)+[1-P(t₁₃<4.23)]= 0.000492 + (1-0.999508)= 2*0.000492= 0.000984≅ 0.001
The p-value: 0.001 is less than α: 0.05, the decision is to reject the null hypothesis.
I hope it helps!