An automobile manufacturer claims that its jeep has a 31.2 miles/gallon (MPG) rating. An independent testing firm has been contracted to test the MPG for this jeep since it is believed that the jeep has an incorrect manufacturer's MPG rating. After testing 230 jeeps, they found a mean MPG of 31.4. Assume the standard deviation is known to be 2.5. A level of significance of 0.05 will be used. Make a decision to reject or fail to reject the null hypothes

we will fail to reject the null hypothesis and conclude that there is insufficient evidence to support the claim that the jeep has an incorrect manufacturer's MPG rating

Step-by-step explanation:

We are given;

Population mean; μ = 31.2

Sample mean; x¯ = 31.4

Sample size; n = 230

standard deviation; σ = 2.5

Significance level = 0.05

Let's define the hypotheses;

Null hypothesis; H0: μ = 31.2

Alternative hypothesis; Ha: μ ≠ 31.2

Formula for the test statistic is;

z = (x¯ - μ)/(σ/√n)

z = (31.4 - 31.2)/(2.5/√230)

z = 0.2/0.4564

z = 0.44

From online p-value from z-score calculator attached and using;

z = 0.44; two tailed hypothesis; significance value = 0.05

We have;

P-value = 0.659937

This p-value is greater than the significance value and thus, we will fail to reject the null hypothesis and conclude that there is insufficient evidence to support the claim that the jeep has an incorrect manufacturer's MPG rating

the answer is 0.000652422

but if you ment to ask what is 6131 divided by 4, it is 1532.75.

step-by-step explanation: