Question

A study was conducted on the effect of an oral antiplaque rinse on plaque buildup on teeth. Twelve people whose teeth were thoroughly cleaned and polished were randomly assigned to two groups of six subjects each. Both groups were assigned to use oral rinses (no brushing) for a two-week period. Group 1, the control group, received a rinse that, unknown to the subjects, contained no antiplaque agent. Group 2 used a rinse that contained an antiplaque agent. A plaque index x, a measure of plaque buildup, was recorded at 4, 7, and 14 days. The mean and standard deviation for the 14-day plaque measurements are shown in the table for the two groups. Control Group Antiplaque Group Sample size Mean 1.29 0.72 Standard deviation 0.33 0.33 (a) State the null and alternative hypotheses that should be used to test the effectiveness of the antiplaque oral rinse. Ho: (- 42) = 0 versus H: (u - 42) = 0 O Ho: (u– uz) < Oversus H : (u – uz) > 0 O Ho: (uuz) = 0 versus H : (u – uz) <0 O Ho: (41 - 42) = 0 versus H: (41 - 42) > 0 O Ho: (41 - 42) # O versus H : (41 - 2) = 0 (b) Do the data provide sufficient evidence to indicate that the oral antiplaque rinse is effective? Test using a = 0.05. (Round your answers to three decimal places.) Test statistic: t = Rejection region: If the test is one-tailed, enter NONE for the unused region. Conclusion: O He is not rejected. There is sufficient evidence to indicate that the oral antiplaque rise is effective. O He is not rejected. There is insufficient evidence to indicate that the oral antiplaque rise is effective. O He is rejected. There is insufficient evidence to indicate that the oral antiplaque rise is effective. Ho is rejected. There is sufficient evidence to indicate that the oral antiplaque rise is effective. (c) Find the approximate p-value for the test. O p-value < 0.005 O 0.005< p-value < 0.010 0.010 < p-value < 0.025 0.025 < p-value < 0.050 0.050 <p-value < 0.100 p-value > 0.100

Answer 1

Solution e study was conducted on the effect of on: Required information is given by Wampkisize mean 6 1.29 0.83 6 0-12 0.33 Glancord deviation Calculate :- @ Slate the null and alternative 1470 therick he hale to obtained the ball and alternative hypothesis (fere, suell hypothesis is [Ho: e11-02 =o) ond, alternative hypothesis is THA: eri-11230] Therefore, option ☺ o correct answer - cto: eu-l2 =0 llorsus Hilli-l2>d =

Do the data Gof friends teke hala to polained the last Slalithe, any t critic Calia is Gere test statistic is 6.29 -0.72] = 0.21)? 0032 5 + v@[16:293) 0:57 Ver [0.013153 = (0.51] 0.0063 - 10:57 = 10052 Lo. 1905 = 2.9917€ 2.992 Therefore, test Statistic is 2.992 + critical value is)

teritical at [Qid=nith:-)) = t[0.05, df =6+6-)) = 460.05, df=12-2) =100.05)10) It-Critico 181241 Dection Begion ! to compared t-test and t-critical value is we have (fere, t-test Q-992 > Hcritical 1.8124 Therefore, option © is correct answer > Rejected to a+Q=0.05) @i) Conclusion We have to conclude that the given Salation it, Here, > Ho its revected. There is a Culficient evidence to indicate that the dental antiplazie wise its effective Therefore, option is correct answer

rini anpriale p value! hale hate to oblained the plaluia for the height 19) planla (+- hout, dif] - t2.999, 10) p-view = 0.006 15 ipvallue = 0.00687 -ਟਾਚਾ , picalcuno <0005 -b option 1 ils Conecte answer