Question

Suppose that consumers test 21 hybrid sedans and get a mean of 30 mpgwith a standard deviation of 7 mpg. Thirty-one non-hybrid sedans get a mean of 21 mpg with a standard deviation of three mpg. Suppose that the population standard deviations are known to be six and three, respectively.

What is the test statistic?

What is the P-Value?

State the distribution to use for the test

Answer 1

Here we need to test the two variables i.e. 1 is non-hybrid sedans and another is hybrid sedans.

Let's consider the  µ1 is the mean of hybrid sedans and µ2 is mean of non hybrid sedans.

Here we need to check the following hypothesis :

. H0: ?1 ? ?2 VS

Ha: ?1 < ?2?

distribution to use for the test is Normal.

So here we need to use the Z-test for Two Means as population standard deviations are known

so the formula for a z-statistic for two population means is:

z =

(x?₁-x?₂ ) /   ? ( ?²₁/n₁) + ( ?²₂/n₂)

The null hypothesis is rejected when the z-statistic lies on the rejection region, which is determined by the significance level (alpha) and the type of tail (two-tailed, left-tailed or right-tailed).

The provided sample means are shown below:

x? 1 = 30

x? 2 = 21

Also, the provided population standard deviations are:

?1 = 6 ?2?=3

and the sample sizes are n1 = 21 n2 = 31

The z-statistic is computed as 6.357.

Using the P-value approach: The p-value is p = 0p=0, and since p = 0 < 0.05p=0<0.05, it is concluded that the null hypothesis is rejected.

At the 5% significance level, there is sufficient evidence to conclude that the mean miles per gallon of non-hybrid sedans is less than that of hybrid sedans