An automobile manufacturer claims that its car has a 35.935.9 miles/gallon (MPG) rating. An independent testing firm has been contracted to test the MPG for this car since it is believed that the car has an incorrect manufacturer's MPG rating. After testing 220220 cars, they found a mean MPG of 35.635.6. Assume the standard deviation is known to be 2.22.2. A level of significance of 0.020.02 will be used. Make a decision to reject or fail to reject the null hypothesis.

Answer with explanation:Let [tex]\mu[/tex] be the population mean.For the given claim , we haveNull hypothesis : [tex]H_0: \mu=35.9[/tex]Alternative hypothesis :  [tex]H_a: \mu\neq35.9[/tex]Since alternative hypothesis is two-tailed , so the test is a two-tailed test.Given : Sample size : n=220  ;Sample mean: [tex]\overline{x}=35.6[/tex]  ;Standard deviation: [tex]s=2.2[/tex]Test statistic for population mean:[tex]z=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}[/tex][tex]z=\dfrac{35.6-35.9}{\dfrac{2.2}{\sqrt{220}}}\approx-2.02[/tex]By using the standard normal distribution table of z , we haveP-value ( two tailed test ) : [tex]2P(Z>|z|)=2(1-P(Z<|z|))[/tex][tex]=2(1-P(z<|-2.02|))=2(1-0.9783083)=0.0433834[/tex]Since , the P-value is greater than the significance level of [tex]\alpha=0.02[/tex] , it means we do not have sufficient evidence to reject the null hypothesis.