Question: The times to failure of certain electronic components in accelerate
The 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 from an exponential population, use the results of Exercise 12.21 and Theorem 12.2 to test the null hypothesis θ = 15 minutes against the alternative hypothesis θ ≠ 15 minutes at the 0.05 level of significance.(Use ln 1.763 = 0.570.)
Answer to relevant QuestionsA single observation of a random variable having a geometric distribution is used to test the null hypothesis θ = θ0 against the alternative hypothesis θ = θ1 > θ0. If the null hypothesis is rejected if and only if the ...Given a random sample of size n from a normal population with the known variance σ2, show that the null hypothesis µ = µ0 can be tested against the alternative hypothesis µ ≠ µ0 with the use of a one- tailed criterion ...Show that the following computing formula for x2 is equivalent to the formula on page 372: Test at the 0.05 level of significance whether the mean of a random sample of size n = 16 is “significantly less than 10” if the distribution from which the sample was taken is normal, = 8.4, and σ = 3.2. What are the ...With reference to Example 13.4, for what values of x1 – x2 would the null hypothesis have been rejected? Also find the probabilities of type II errors with the given criterion if (a) µ1 – µ2 = 0.12; (b) µ1 – µ2 = ...
Post your question