Question

44%E), , Tue 1:24 PM Q ! Return to Blackboard y Statistics, 9e Help l e Read, Study & Practice Assignment nmentOpen Assignment NMENT RESOURCES Chapter 09, Section 9.3, Problem 037 h-nd Chaeter 09. A random sample of 17 observations taken from a population that is n ally distributed produced a sample mean of 42.4 and a standard deviation of 8. Find the range for the p-value and the critical and observed values of t for each of the following tests of hypotheses using, ao Use the t distribution table to find a range for the p-value. Round your answers for the values of t to three decimal places a.Ho: ? = 46 versus H1:?<46. ed Chapter 09, enter 09 Section tProblem 006 apter 02 Section p-value < Probem 019 aeter 09, Section apter 09, Section Probiem 029a b. Ho: H 46 versus : 46. . Probiem 042 < p-value < -ed Chanter 09 critical let tcritical right- tobserved aeter 09, Section Probiem 036b apter 09. Section Question Attempts: 0 of 3 used sAvs sOR LATER

Answer 1

Solution:

a)

The provided sample mean is X = 42.4 and the sample standard deviation is s = 8, and the sample size is n 17 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested Ho: μ-46 This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is α = 0.01, and the critical value for a left- tailed test is te2.583. The rejection region for this left-tailed test is R-t : 3) Test Statistics The t-statistic is computed as follows 2.583 X s/v/n 42.4-46 = 8/v_ = t = 1.855 Since it is observed that t1.855 is not rejected. te2.583, it is then concluded that the null hypothesis Using the P-value approach: The p-value is p0.041, and since p concluded that the null hypothesis is not rejected. 0.041 0.01, it is

b)

The provided sample mean is X = 42.4 and the sample standard deviation is s = 8, and the sample size is n 17 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested Ho: μ-46 Ha: 46 This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is α-0.01, and the critical value for a two-tailed test is t-2.921 The rejection region for this two-tailed test is R 3) Test Statistics The t-statistic is computed as follows {t: t>2.921) X-140 42.4 46 t = 1.855 (4) Decision about the null hypothesis Since it is observed that t-1.855 < te-2.921, İt is then concluded that the null hypothesis is not rejected. Using the P-value approach. The p-value is p = 0.0821, and since p = 0.0821 > 0.01, it is concluded that the null hypothesis is not rejected.

Done