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The data found below measure the amounts of greenhouse gas emissions from three types of vehicles....

The data found below measure the amounts of greenhouse gas emissions from three types of vehicles. The measurements are in tons per​ year, expressed as CO2 equivalents. Use a 0.025 significance level to test the claim that the different types of vehicle have the same mean amount of greenhouse gas emissions. Based on the​ results, does the type of vehicle appear to affect the amount of greenhouse gas​ emissions? LOADING... Click the icon to view the data. Type A 6.9 6.9 7.4 6.9 7.6 6.9 6.5 7.7 6.9 6.3 Type B 8.1 7.2 7.3 7.6 8.8 9.1 8.8 8.7, Type C 8.4 9.7 8.3 9.9 8.9 9.2 9.8 9.3 9.1

I need the the std. test statistic and explanation on how to come up with this number and critical values plus the determination on whether to reject or not to reject the hypothesis

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Answer #1

There is a lot of information on this test

Search F test on Google and you will get.

If you know about Expected value in statistics,

Then I will add one more thing:

Under H0, Expected value of numerator and denominator in the statistic is supposed to be (sigma)^2. So if H0 is true, the value of the test statistic will be small and hence we will not reject the null hypothesis and claim that it is true.

When to use the F test?

When you will find that you want to test the difference in effects of the same factor,

Suppose you have factor 1: vehicle of type 1

and factor 2: coal of type 1

Then they cannot be tested by ANOVA

But when you are having

Factor 1: vehicle 1

......

Factor n: vehicle n

Then you can do one way ANOVA to test whether all the vehicles type have same of different effects.

We will basically test whether the means of each group are different or same.(Means because we can compare the means of each group)

What is the critical value?

On getting the value of F Statistic, we need to compare it with a value to see whether to accept or reject the null hypothesis.

Now at 0.025 level of significance,

We have under H0

P(F (2,21)>k) = 0.025

So k= F (0.025,2,21)

So k is the critical value above which we will reject the null hypothesis.

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