Question

In an experiment to compare the tensile strengths of I = 5 different types of copper wire, J = 4 samples of each type were used. The between samples and within samples estimates of σ2 were computes as MSTr = 2673.3 and MSE = 1094.2, respectively.
a) Use the F test at level 0.05 to test H0 : μ1 = μ2 = μ3 = μ4 = μ5 versus Ha : at least two μi 's are unequal.

b) What can be said about the P-value for the test?

Answer 1

Answer:

Step-by-step explanation:

Hello!

You need to test if the population means of 5 populations are equal. To do so, with the given hypothesis, the best is to conduct an ANOVA. The hypothesis test for the ANOVA is a one-tailed F-test.

a) Hypothesis:

H₀: μ₁= μ₂= μ₃= μ₄= μ₅

H₁: At least two μi are unequal.

α: 0.05

The statistic is:

$F= \frac{MSTr}{MSE}$~$F_{I_1-1;J_2-1}$

As I said before, this is a one-tailed test, which means that you will reject the null hypothesis when the values of the statistic are big.

Critical value:

$F_{4;3;0.95} = 9.12$

If F ≥ 9.12, then reject the null hypothesis.

If F < 9.12, then do not reject the null hypothesis.

$F_{H0}= \frac{2673.3}{1094.2}= 2.44$

Since the calculated value is less than the critical value, the decision is to not reject the null hypothesis. This means that the population means are equal.

b) The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).

You calculate it as:

P(F≥2.44) = 1 - P(F<2.44) = 1 - 0.755203 = 0.244797

As you see the p-value is greater than the level of significance, so with this method you reach also decide to not reject the null hypothesis.

I hope you have a SUPER day!