true or false: When testing H0: u = – 8 against H1: u > – 8, the power of the test is greater for μ1 = – 6 than for μ1 = – 7, all other things being the same
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true or false: When testing H0: u = – 8 against H1: u > – 8, the power of the test is greater for μ1 = – 6 than for μ1 = – 7, all other things being the same.
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- 4. The lifetime, in years, of a type of small electric motor operating under adverse conditions is T ~Γ(6, 3.6). Whenever the motor fails, it is replaced with another of the same type. Find theprobability that fewer than 6 motors fail within one year.You are using t-test with 15 degrees of freedom to test H0: u=112 vs Ha: u<112, and the rejection was defined to be all the values less than -1.753. What is the implied value of alpha that this test is being performed at?Is there a way to solve this w/o using L'Hospital's rule?
- If after performing a t-test for comparison of means (alpha= .05) we obtain p=0.0256, what is our conclusion? a. Fail to reject H0 b. Reject H0 c. Reject H1 d. Unable to determine from the information provided1. A T interval for the difference of 2 population means was obtained on the TI, as shown below. What can be concluded?A test of H0H0:μ=1:μ=1 against HaHa:μ>1:μ>1 has test statistic zz = 1.74. Answer "Yes/Y" or "No/N" to the following questions. Is this test significant at the 2.5% level (αα = 0.025)? Is it significant at the 0.5% level (αα = 0.005)?
- true or false. if you found X2 = 10 with df= 5 you would fail to reject Ho at the 5% significance level.Dr. Chapman conducted an experiment in which participants watched paint dry for 30 minutes twice, once being paid $1 and once being paid $30. When comparing the samples, he calculated t = 3.57. He assumed α = .01 with df = 5, so the tcv = ±4.032. Because the calculated value was: A. greater than the critical value, Dr. Chapman can reject the null hypothesis. B. greater than the critical value, Dr. Chapman failed to reject the null hypothesis. C. less than the critical value, Dr. Chapman failed to reject the null hypothesis. D. less than the critical value, Dr. Chapman can reject the null hypothesis.See the attached image for the introduction. In terms of variables xi and parameters βi, write the null and alternative hypotheses for testing whether, after including Price/Square Feet(x2) in the model already, the further incorporation of the other 2 explanatory variables (x1, x3) adds any useful information for explaining pricey. Also, give the value of the F statistic and its degrees of freedom (df).
- A regression on the original regressors, ?̂t2 and a constant term yields the following statistics: R2 = 0.296041 F = 1.177507 coeff of ?̂t2 has a t-statistic of 2.876 With this information, which test can you implement to deal with the problem omitted variables and why? Implement the test as stated in b(i) and interpret the results. What is (are) the consequence(s) of the problem alluded to above on the estimators?The test statistic of z=−2.97 is obtained when testing the claim that p=3/5. a. Using a significance level of a=0.10, find the critical value(s). b. Should we reject H 0or should we fail to reject H 0?Persons having Raynaud's syndrome are apt to suffer a sudden impairment of blood circulation in fingers and toes. In an experiment to study the extent of this impairment, each subject immersed a forefinger in water and the resulting heat output (cal/cm2/min) was measured. For m = 9 subjects with the syndrome, the average heat output was x = 0.65, and for n = 9 nonsufferers, the average output was 2.03. Let μ1 and μ2 denote the true average heat outputs for the sufferers and nonsufferers, respectively. Assume that the two distributions of heat output are normal wit