Question

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 When bonding teeth, new bonding adhesive is not as strong as the current standard, a composite adhesive. Tests on a sample of 8 extracted teeth bonded with the new adhesive resulted in a mean breaking strength (after 24 hours of x=5.35 Mpa and a standard deviation of s=0.39 Mpa. Orthodontists want to know if the true mean breaking strength is less than 5.94 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. B. Ho : μ巧.94 vs. Ha : μ#5.94 D. Ho : μ 5.35 vs. Ha : μ< 5.35 F. Ho : μ= 5.94 vs. Ha : μ< 5.94 A. H0με 5.94 vs. Ha: μ> 5.94 C. Ho : μ: 5.35 vs. Ha: μ# 5.35 E. Ho : μ2 5.94 vs. Ha: μ< 5.94 b. Find the rejection region for the test using α=0.01. Choose the correct answer below. O A. t 3.499 ○ B. D. t>2.998 t#3.499 C. t<-2.998 OE. t>-2.998 O G. t2.988 c. Compute the test statistic. tRound to two decimal places as needed.) d. Give the appropriate conclusion for the test. Choose the correct answer below. F. -3.499 H. t<-3.499 A. Do not reject Ho . There is sufficient evidence to indicate μく5.94 Mpa. O B. Reject Ho . There is insufficient evidence to indicate μ< 5.94 Mpa. Do not reject Ho. There is insufficient evidence to indicate μ< 5.94 Mpa. D. C. Reject Ho . There is sufficient evidence to indicate μ< 5.94 Mpa. e. What conditions are required for the test results to be valid? 0 A. We must assume that the sample was not random and selected from a population 0 B. We must assume that the sample was random and selected from a population with with a highly skewed distribution. We must assume that the sample was not random and selected from a normaly distributed population. a distribution that is approximately normal. C. D. We must assume that the sample was random and selected from a population that is highly skewed.

Answer 1

H_0: mu greaterthanorequalto 5.94 H_a: mu < 5.94 Given X^bar = 5.35, and s = 0.39, n = 8 alpha = 0.01 t_stat = X^bar - mu/(s/ squareroot n) = 5.35 - 5.94/(0.39/ squareroot 8) t_stat = -4.279 -4.28 partialdifferential f = n - 1 = 8 - 1 = 7 therefore t_critical value (left tailed) = -2.998 (c) t < -2.998 Because < t_stat <ta, reject null (D) Reject H_0, there is sufficient Evidence to Indicate mu < 5.94 option B