To compare the variations in weight of four breeds of dogs, researchers took independent random samples of sizes n1 = 8, n2 = 10, n3 = 6, and n4 = 8, and got σ21 = 16, σ22 = 25, σ23 = 12, and σ24 = 24. Assuming that the populations sampled are normal, use the formula of part (b) of Exercise 12.25 to calculate –2 · ln λ and test the null hypothesis σ21 = σ22 = σ23 = σ24 at the 0.05 level of significance. Explain why the number of degrees of freedom for this approximate chi-square test is 3.
Answer to relevant QuestionsThe times to failure of certain electronic components in accelerate environment tests are 15, 28, 3, 12, 42, 19, 20, 2, 25, 30, 62, 12, 18, 16, 44, 65, 33, 51, 4, and 28 minutes. Looking upon these data as a random sample ...Let X1 and X2 constitute a random sample of size 2 from the population given by If the critical region x1x2 ≥ 3/4 is used to test the null hypothesis θ = 1 against the alternative hypothesis θ = 2, what is the power of ...Show that the rule on page 372 for calculating the expected cell frequencies applies also when we test the null hypothesis that we are sampling r populations with identical multinomial distributions. With reference to Example 13.3, use suitable statistical software to find the P-value that corresponds to t = –0.49, where t is a value of a random variable having the t distribution with 4 degrees of freedom. Use this ...An epidemiologist is trying to discover the cause of a certain kind of cancer. He studies a group of 10,000 people for five years, measuring 48 different “factors” involving eating habits, drinking habits, smoking, ...
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