Answer the questions on separate paper. Problems 6 - 7 are take- problems, bring

Question

Answer the questions on separate paper. Problems 6 - 7 are take- problems, bring them back on Thursday, April 21. Consider the test of H_0 : mu = 6.0 versus H_a : mu 6.0. Suppose alpha = 0.05, n = 30 and s = 1.2. Suppose the z-test is used, where s is substituted for sigma. What is the upper rejection region for Y? Assume that sigma = 1.2. Compute the power of the z-test if mu_a = 6.5. Suppose the t-test is used. For what t-values will the null hypothesis be rejected? (two-sided test.) If s = 1.2, for what values of Y will the null hypothesis be rejected? Assume that both s = 1.2 and sigma = 1.2. Compute the power of the t-test if mu_a = 6.5. The waiting time between tremors at a seismological monitoring station is assumed to be Exponentially distributed with mean theta (in hours). A single observation will be used to test the null hypothesis theta = 10 against the alternative theta 10. The null hypothesis will be rejected if the observed waiting time is less than 8 hours or if the observed waiting time is more than 12 hours. Compute the probability of Type I error for this test. What is the probability of Type II error if theta_a = 4? The weights of Black Angus steers at age 12 months are approximately Normally distributed. A simple random sample of 24 steers were weighed, with standard deviation 238 pounds. Test the null hypothesis sigma = 250 against sigma 250 at the 0.01 significance level. (use the confidence interval p. 435.) The manufacturer of a spot remover claims that his product removes 90% of all spots. In a random sample, only 174 of 200 spots were removed with the product. Comment on the manufacturer's claim at the alpha = 0.05 significance level.

Explanation / Answer

State hypotheses:

H0:Pu<0.90

H1:Pu>=0.90

Sampling distribution:Z (large sample)

alpha=0.05.

Test statistic:

Z=(Ps-Pu)/sqrt [Pu(1-Pu)/N], Where Ps and Pu are sample proportion and population proportion, N is sample size.

=(174/200-0.90)/sqrt [0.90(1-0.90)/200]

=-1.41

p value is 0.07927.

P value is not less than alpha=0.05. Therefore, fail to reject null hypothesis. There is not sufficient sample evidence to conclude that manufacturer's claim is right.

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