Help with these from my chapter 22 math lesson, distribution table uploaded

Assume that both populations are normally distributed. Population 1 Population 2 (a) Test whether 1 # H2 at the a = 0.01 level of significance for the n 10 10 given sample data. X 11 9.7 (b) Construct a 99% confidence interval about 1 - H2. S 3 3.2 (a) Test whether 1 # 2 at the a = 0.01 level of significance for the given sample data. Determine the null and alternative hypothesis for this test. O A. HO:My = H2 H1: p1 > H2 OB. HO:My # H2 H1: H1 > H2 OC. HO:My # H2 Hy:My = H2 D. HO:H1 = H2 Hy:My # H2 Detemine the P-value for this hypothesis test. P = (Round to three decimal places as needed.)Use the given statistics to complete parts (a) and (b). Assume that the populations are normally distributed. (a) Test whether M > M at the a = 0.05 level of signicance for the given sample data. (b) Construct a 95% condence interval about (3) Identify the null and alternative hypotheses for this test. 'VA- Ho=i11=112 3- H1=H1>P2 D. H02p1p2 E. H1=H1=P2 Find the test statistic for this hypothesis test. B (Round to two decimal places as needed.) i11'l12- H01P1=u2 H11P1H2 H11P1=P2 MINI: Population 1 24 47.5 6.1 - Hoiil1=i12 H13P1l12 H01P1H2 H1 : My > H 2 Determine the P-value for this hypothesis test. P-value = (Round to three decimal places as needed.)Two researchers conducted a study in which two groups of students were asked to answer 42 trivia questions from a board game. The students in group 1 were asked to spend 5 minutes thinking about what it would mean to be a professor, while the students in group 2 were asked to think about soccer hooligans. These pretest thoughts are a form of priming. The 200 students in group 1 had a mean score of 22.6 with a standard deviation of 3.4, while the 200 students in group 2 had a mean score of 15.7 with a standard deviation of 2.5. Complete parts (a) and (b) below. (a) Determine the 90% condence interval for the difference in scores, u1 - p2. Interpret the interval. The lower bound is The upper bound is (Round to three decimal places as needed.) Ramp metering is a trafc engineering idea that requires cars entering a freeway to stop for a certain period of time before joining the traffic ow. The theory is that ramp metering controls the number of cars on the freeway and the number of cars accessing the freeway, resulting in a freer flow of cars, which ultimately results in faster travel times. To test whether ramp metering is effective in reducing travel times, engineers conducted an experiment in which a section of freeway had ramp meters installed on the on-ramps. The response variable for the study was speed of the vehicles. A random sample of 15 cars on the highway for a Monday at 6 pm. with the ramp meters on and a second random sample of 15 cars on a different Monday at 6 pm. with the meters off resulted in the following speeds (in miles per hour). a Click the icon to view the data sets. (a) Draw side-byside boxplots of each data set. Does there appear to be a difference in the speeds? Are there any outliers? Choose the correct box plot below. o A. o a. o c. 0n orr or or- 9' 0n- 9* 0n 0' mmwrm WW 0 15 30 45 60 E.) 0 15 30 45 60 E 0 15 30 45 60 E MPH MPH MPH Speed data (in MPH) Ramp Meters 0n Ramp Meters Off FU\" Data Set '3. 28 47 57 23 26 41 39 31 26 34 36 32 43 46 50 47 37 19 36 54 41 29 22 39 42 25 48 38 51 40 A researcher wanted to determine if carpeted rooms contain more bacteria than uncarpeted rooms. The table shows the Full data set results for the number of bacteria per cubic foot for both types of rooms. Carpeted Uncarpeted 14.6 8'6 15 12.5 4.9 4.2 9.9 11.8 7.4 4.6 13.7 4.1 8.9 8.6 6.2 4.8 Determine whether carpeted rooms have more bacteria than uncarpeted rooms at the a = 0.05 level of significance. Normal probability plots indicate that the data are approximately normal and boxplots indicate that there are no outliers. State the null and alternative hypotheses. Let population 1 be carpeted rooms and population 2 be uncarpeted rooms. O A. HO: H1 = H2 Hy: My # H2 OB. HO: My

H2 C. HO: H1 = H2 H1: H1 > H2 OD. HO: H1 = H2 Hy: My H2 C. Ho: H1 = H2; Ha: My