Answer Happy

When bonding to orthodontistain a dryf Anew bonding adhesive has been developed to me theme of dry However there is not the windows Standard, a composite thesive Tests on a sample of 11 extraded to be with the adherested in a man breaking Status Mpenda standards Media to know the true mean breaking this than 15 Meman being strength of the computer C 2. Set up the noland temative hypotheses for the best chose conectar below OAM 515H616 OCH 515 515 OLH 6) vs 43 On 557515 ODHOH OR #515 2515 Find the rejection for the testing 005 Choose the connect we below O 1812 OD-2221 OF 2 OH)2228 OA1112 OC2220 OT-1012 OG 1812 Compute the cando decimal places as needed d. Give the appropriate conclusion for the last Choose the comedianweb DA Reece There iniciantvidence to indicate <515 Mpa O Do not reject There is an evidence to indicate <515 Mpa OC Red Theresent evidence to indicate <515 M O. Do not reject H, There is sufficient evidence to indicate < 515 Ma What contions are equired for the loss to be valin OA We must assume that the sample was random and selected from a population with a distribution that is approximately normal OC. We must asume that the sample was not random and selected from anomaly distributed population on. We must me that the same was not random and droppo dhton Oo. We must me that the sample random and select the population that
When bonding teeth, orthodontists must maintain a dry field. A new bonding adhesive has been developed to eliminate the necessity of a dry field. However, there is concern that the new bonding adhesive is not as strong as the current standard, a composite adhesive Tests on a sample of 11 extracted teeth bonded with the new adhesive resulted in a mean breaking strength (after 24 hours) of x = 4 63 Mpa and a standard deviation of s = 0.34 Mpa. Orthodontists want to know if the true mean breaking strength is less than 5.15 Mpa, the mean breaking strength of the composite adhesive. a. Set up the null and alternative hypotheses for the test Choose the correct answer below. O A. Ho H=515 vs. H. H> 5.15 OC. Ho 25.15 vs. Ha u<5.15 O E. Ho - 463 vs. H. H<463 OB. Ho H=5,15 vs H, »#515 OD. Ho u=463 vs H, 174.63 OF. Ho H=5.15 vs. H. <5.15 b. Find the rejection region for the test using a=0 05 Choose the correct answer below. O A. 1.812 od -2.228 O E t - 1.812 O G. > 1.812 OB. t< - 1.812 OD. > -2.228 OF. t<2.228 OH. > 2.228 C. Compute the test statistic t EH (Round to two decimal places as needed) d. Give the appropriate conclusion for the test. Choose the correct answer below. O A. Reject Ho There is insufficient evidence to indicate u <5.15 Mpa O B. Do not reject Ho There is insufficient evidence to indicate p < 5.15 Mpa. OC. Reject Ho There is sufficient evidence to indicate u <5.15 Mpa OD. Do not reject Ho There is sufficient evidence to indicate u <5 15 Mpa.
e. What conditions are required for the test results to be valid? O A. We must assume that the sample was random and OB. We must assume that the sample was not random selected from a population with a distribution that is and selected from a population with a highly skewed approximately normal distribution O C. We must assume that the sample was not random OD. We must assume that the sample was random and and selected from a normally distributed population selected from a population that is highly skewed.