**Does Octane Matter?**Some people believe that different octane gasoline results in different miles per gallon in a vehicle. Below is a sample of 11 cars. Each person was asked to drive their car for as long as a 10-gallon tank would take them. The mileage was recorded as follows:| Person | Miles on 87 Octane | Miles on 92 Octane ||--------|--------------------|--------------------|| 1 | 234 | 237 || 2 | 257 | 238 || 3 | 243 | 229 || 4 | 215 | 224 || 5 | 114 | 119 || 6 | 287 | 297 || 7 | 315 | 351 || 8 | 229 | 241 || 9 | 192 | 186 || 10 | 204 | 209 || 11 | 547 | 562 |**Research Question:**Do the data support that different octanes produce different miles per gallon at the \(\alpha = 0.02\) level of significance?**Tasks:**a) Express the null and alternative hypotheses in symbolic form for this claim. Assume \(d = \) (mileage on 92 octane - mileage on 87 octane). Use \( \neq \) for not equal.- \( H_0: \mu_d = 0 \)- \( H_a: \mu_d \neq 0 \)b) What is the significance level?- \( \alpha = 0.02 \)c) What is the test statistic? Round to 3 decimal places.- Test statistic \( = \) [Provide space for answer]d) What is the p-value? Round to 5 decimal places.- p = [Provide space for answer]

Solution: Let x= Miles on 87 octane and y =Miles on 92 octane x y d =y-x (d-d) (d-d)2 234 237 3 -2.0909 4.371863 257 238 -19 -24.0909 580.3715 243 229 -14 -19.0909 364.4625 215 224 9 3.9091 15.28106 114 119 5 -0.0909 0.008263 287 297 10 4.9091 24.09926 315 351 36 30.9091 955.3725 229 241 12 6.9091 47.73566 192 186 -6 -11.0909 123.0081 204 209 5 -0.0909 0.008263 547 562 15 9.9091 98.19026 n= 11 Sample size ∑d =56d=∑dn=5611=5.0909 Difference of sample mean ∑(d-d)2=2212.909 sd= ∑(d-d)2n-1sd= 2212.90910sd= 221.2909sd= 14.8759 Difference of sample standard deviation a) The null and alternative hypothesis is Ho:μd= 0 VS Ha:μd≠0 b) The level of significance is α=0.02 c) Test statistic t=dsd/nt=5.090914.8759/11t=5.090914.8759/3.316625t=5.09094.485253=1.135031t= 1.135 Test statistic t= 1.135 d) To find p value p value = 2×P(t>1.135) p value = 0.28285 From Excel The p value for the test is 0.28285