Question

A consumer group plans to test whether a new passenger car that is advertised to have a mean highway miles per gallon of at least 33 actually meets this level. They plan to test the hypothesis using a significance level of 0.05 and a sample size of n = 100 cars. It is believed that the population standard deviation is 3 mpg. Based upon this information, if the "true" population mean is 32.0 mpg, what is the probability that the test will lead the consumer group to "accept" the claimed mileage for this car? Question 36 options: About 0.9545 Approximately 0.0455 About 0.45 None of the above

Answer 1

Probability that the test will lead the consumer group to "accept" the claimed mileage for this car is 0.00043.

Explanation:

We are given that a consumer group plans to test whether a new passenger car that is advertised to have a mean highway miles per gallon of at least 33 actually meets this level.

It is believed that the population standard deviation is 3 mpg. Based upon this information, the "true" population mean is 32.0 mpg.

Let
`\bar X` = sample mean highway miles per gallon

The z-score probability distribution for sample mean is given by;

Z =
`(\bar X -\mu)/((\sigma)/(√(n) ) )` ~ N(0,1)

where,
`\mu` = true population mean = 32.0 mpg

`\sigma` = population standard deviation = 3 mpg

n = sample of cars = 100

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that a mean highway miles per gallon of at least 33 actually meets this level is given by = P(
`\bar X`
`\geq` 33)

P(
`\bar X`
`\geq` 33) = P(
`(\bar X -\mu)/((\sigma)/(√(n) ) )`
`\geq`
`(33-32)/((3)/(√(100) ) )` ) = P(Z
`\geq` 3.33) = 1 - P(Z < 3.33)

= 1 - 0.99957 = 0.00043

The above probability is calculated by looking at the value of x = 3.33 in the z table which has an area of 0.7673.

Therefore, the probability that the test will lead the consumer group to "accept" the claimed mileage for this car is 0.00043.