87-octane 234 258 242 215 113 288 314 229 191 205 547 92-octane 238 238 229 225 119 298 351 241 187 210 561 Expected z-score Expected Z-score Expected z-score -200 700 75 87 Octane 92 Octane Difference Paired T-Test and CI: 92 Octane, 87 Octane Paired T for 92 Octane - 87 Octane Z Mean StDev SE Mean 92 Octane 11 263.364 114.641 34.566 87 Octane 11 257.818 109.271 32.946 Difference 11 5.545 14.942 4.505 T-Test of mean difference =0 (vs > 0): T-Value = 1.23 P-Value =0.123Critical Values for Use of the Correlation Coefficient with Normal Probability Plots Sample Size, n Critical Value Sample Size, n Critical Value 5 0.880 16 0.941 6 0.888 17 0.944 7 0.898 18 0.946 8 0.906 19 0.949 9 0.912 20 0.951 10 0.918 21 0.952 11 0.923 22 0.954 12 0.928 23 0.956 13 0.932 24 0.957 14 0.935 25 0.959 15 0.939 30 0.960Some people believe that higher-octane fuels result in better gas mileage for their car. To test this claim, a researcher randomly selected 11 individuals to participate in the study. Each participant received 10 gallons of gas and drove his car on a closed course. The number of miles driven until the car ran out of gas was recorded. A coin flip was used to determine whether the car was filled up with 87-octane or 92-octane first, and the driver did not know which fuel was in the tank. Complete parts (a) through (e). Click here to view the data. probability plots, and technology output. Click here to view the table of critical values for the correlation coefficient. (a) Why is it important that the matching be done by driver and car? O A. It allows each driver to determine which fuel is best for their car. O B. How someone drives and the car they drive result in different fuel consumption. O C. It allows all of the trials to be done at the same time. O D. It cuts the cost of doing the research (b) Why is it important to conduct the study on a closed track? O A. So that the researcher can watch the drivers O B. So that each car uses the same amount of gas O C. So that all drivers and cars have similar driving conditions O D. So that each car travels the same distance (c) The normal probability plots and linear correlation coefficients for miles on 87-octane and miles on 92-octane are given. The correlation between 87 octane and the expected z-scores is 0.877. The correlation between 92 octane and the expected z-scores is 0.879. Are either of these variables normally distributed? O A. Yes, 92-octane is normally distributed. O B. No, neither variable is normally distributed. O C. Yes, 87-octane is normally distributed. O D. Yes, both variables are normally distributed. (d) The differences are computed as 92-octane minus 87-octane. The normal probability plot of the differences is shown. The correlation between the differenced data and the expected z-scores is 0.957. Is there reason to believe that the differences are normally distributed? since the is (e) The researchers used a statistical software package to determine whether the mileage from 92-octane is greater than the mileage from 87-octane. What do you conclude at a = 0.05? Why? Begin by writing the hypotheses. Ho: The difference in mileage for 92-octane and 87-octane zero. H,: The difference in mileage for 92-octane and 87-octane zero. The test statistic is (Do not round.) The P-value is (Do not round.) Therefore, there sufficient evidence that the mileage from 92-octane the mileage from 87-octane