Question

An automobile manufacturer claims that its car has a 42.8 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 300 cars, they found a mean MPG of 42.6. Assume the variance is known to be 5.29. A level of significance of 0.02 will be used. 1. Find the value of the test statistic. Round your answer to 2 decimal places.

Answer 1

Answer: -1.51

Explanation:

Let
`\mu` be the population mean rating of car.

As per given , we have

`H_0: \mu=42.8`

`H_a: \mu\\eq42.8`

Since population variance is known, so we use z-test.

Test statistic :
`z=\frac{\overline{x}-\mu}{\sqrt{(\sigma^2)/(n)}}`

, where n= sample size

`\overline{x}` = Sample mean

`\sigma^2`= variance

As per given ,

n=300

`\overline{x}=42.6`

`\sigma^2=5.29`

`\mu=42.8`

Put these values in formula we get

`z=\frac{42.6-42.8}{\sqrt{(5.29)/(300)}}`

`z=(-0.2)/(√(0.0176333))`

`z=(-0.2)/(0.13279)=-1.50613751035\approx1.51`

Hence, the value of the test statistic is z= -1.51 .