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?

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:

~

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:

If F ≥ 9.12, then reject the null hypothesis.

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

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!

answer: 7270

step-by-step explanation:

homozygous. one is homozygous, while two is heterozygous.