Suppose that an automobile manufacturer advertises that its new hybrid car has a mean gas mileage of 50 miles per gallon. You take a simple random sample of n = 30 hybrid vehicles and test their gas mileage. You find that in this sample, the average is 47 miles per gallon with a standard deviation of 5.5 miles per gallon. Is there enough evidence to support the advertised claim using α = 0.05? What is the null and alternative hypothesis for this problem?

Step-by-step explanation:

Given that,

H0 : µ = 50

Ha : µ ≠ 50

α = 0.05

x¯ = 47

s = 5.5

To find the score, we will use the formula below,

z =(x¯ − µ)/s/√n

z =47 − 50/5.5/√30

z = −3.0/1.0042

z = -2.988

z ≈ -3

The cumulative area for this z-score is 0.0013.

This is a two-tailed test:

P = 2(0.0013) = 0.0026.

Since P = 0.0026 < 0.05 = α, we make the decision to reject the null hypothesis.

This means that there is not enough statistical evidence to support the advertiser’s claim of an average of 50 miles per gallon.

Also,

The critical value for a two tailed test is zc = 1.96

Since it is observed that

|z| = |-3| = 3> 1.96 = zc

It is then concluded that the null hypothesis is rejected.

Confidence interval

Xˉ ± zc • σ/√n

47 ± 1.96 ×5.5/√30

47 ± 1.96

(45.032, 48.968) gallons

x^6

step-by-step explanation:

this is correct because:

x-6=0

0*x=0

so

x^6