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posledela
4 years ago
12

9) 2n - 14=4n + 4n what’s the answer

Mathematics
1 answer:
Artist 52 [7]4 years ago
6 0

The answer is

n = -7/3 or -2.333

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An article in the Journal of Materials Engineering (Vol 11, No. 4, 1989, pp. 275-282) reported the results of an experiment to d
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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

t= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ~~t_{n_1+n_2-2}

Sa= \sqrt{\frac{(n_1-1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2} } = \sqrt{\frac{8*4.28+5*5.62}{9+6-2} }= \sqrt{4.7953}= 2.189= 2.19

t_{H_0}= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } = \frac{(16.36-11.48)-0}{2.19*\sqrt{\frac{1}{9} +\frac{1}{6} } } = 4.23

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!

6 0
3 years ago
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