Problem 5A company is marketing an additive for gasoline to improve car mileage per gallon of gasolineconsumed. Ten cars were selected, and the highway miles per gallon (mpg) for each car withoutand with the additive were measured. Determine whether there is enough statistical evidencethat the mpg is improved upon using the additive with 95% confidence and the p-valueapproach. **Car Fuel Efficiency Study**The table below displays a comparative analysis of the fuel efficiency, measured in miles per gallon (mpg), of ten cars both with and without the use of a fuel additive. This data demonstrates how the additive impacts fuel consumption.| Car | MPG without Additive | MPG with Additive ||-----|----------------------|-------------------|| 1 | 25.5 | 25.0 || 2 | 19.0 | 20.1 || 3 | 18.2 | 19.2 || 4 | 17.5 | 18.2 || 5 | 24.0 | 24.0 || 6 | 21.0 | 21.5 || 7 | 27.5 | 27.4 || 8 | 18.5 | 20.0 || 9 | 19.4 | 20.1 || 10 | 26.0 | 26.2 |**Analysis:**- **General Observation:** In most cases, there is an increase in fuel efficiency with the addition of the additive.- **Significant Changes:** Car 2, Car 3, Car 4, and Car 8 show noticeable improvements in mpg with the additive.- **Minimal or No Change:** Car 5 maintains the same mpg with and without the additive, and changes in Cars 1 and 7 are negligible.This information can be useful for understanding the benefits of using fuel additives in enhancing car performance in terms of fuel consumption.

Given that: Car mpg without additive mpg with additive 1 25.5 25 2 19 20.1 3 18.2 19.2 4 17.5 18.2 5 24 24 6 21 21.5 7 27.5 27.4 8 18.5 20 9 19.4 20.1 10 26 26.2Evaluate Σd and Σd2 by using the following table: mpg without additive mpg with additive d d2 25.5 25 0.5 0.25 19 20.1 -1.1 1.21 18.2 19.2 -1 1 17.5 18.2 -0.7 0.49 24 24 0 0 21 21.5 -0.5 0.25 27.5 27.4 0.1 0.01 18.5 20 -1.5 2.25 19.4 20.1 -0.7 0.49 26 26.2 -0.2 0.04 Σd=-5.1 Σd2=5.99 Compute the mean of difference by using the formula: d¯=Σdn=-5.110=-0.51 Compute the standard deviation of difference by using the formula: sd=Σd2-Σd2nn-1=5.99--5.12109=0.614 State the null and alternative hypotheses: H0:μd≥0H1:μd<0 Obtain the significance level: α=1-Confidence level=1-0.95=0.05 Find the degrees of freedom: df=n-1=10-1=9Compute the test statistic: t=d¯-μdsdn=-0.51-00.61410=-2.63 The value of test statistic is -2.63. Determine the p-value of the test statistic by using the following excel function: Make Decision: Rejection rule: Reject H0 if p-value of the test statistic < significance level, α=0.05, otherwise do not reject it. ∵0.014<0.05 Reject it. Conclusion: There is enough statistical evidence to conclude that the mpg is improved upon using the additive at 0.05 significance level.