In this specific example, $p=0.175$, but you'd need a ruler to calculate that by hand :-) You can tell that the $F$ statistic here would result in a failure to reject the null hypothesis because it is less than $C$, that is, its $p$ value is greater than. The area under the curve from $C$ to infinity equals the significance level (here, 5%). The critical value $C$ also makes an appearance in this diagram. You should notice a couple of other things about the distribution under null hypothesis:ДЎ) $F$ values approaching zero are highly unlikely (this is not always true, but it's true for the curve in this example)Дў) After a certain point, the larger the $F$ is, the less likely it is. Notice that every value of $F$ must correspond to a unique $p$ value, and that higher $F$ values correspond to lower $p$ values.

The $p$ value is the dark blue area under the curve from $F$ to infinity. In this diagram, the observed $F$ statistic is the distance from black dashed line to the vertical axis. The curve here is an $F$ distribution, that is, the distribution of $F$ statistics that we'd see if the null hypothesis were true. The best way to think about the relationship between $F$, $p$, and the critical value is with a picture: For more complicated models the degrees of freedom get more complicated, but follow similar ideas. The numerator degrees of freedom are based on the number of groups that you are comparing (for 1-way it is the number of groups minus 1) and the denominator degrees of freedom are based on the number of observations within the groups (for 1-way it is the number of observations minus the number of groups). The F distribution has 2 types of degrees of freedom. If you would rather use a rejection region than a p-value, then you can use the F table or the qf function in R (or other software). Looking up this ratio on the F-table (or computing it with a function like pf in R) will give the p-value. If the null is false, then the numerator will be large relative to the denominator and the ratio will be greater than 1. So when the null of all means equal is true then the 2 measures (with some extra terms for degrees of freedom) will be similar and the ratio will be close to 1.

The denominator is an average of the sample variances for each group, which is an estimate of the overall population variance (assuming all groups have equal variances). But if the null hypothesis is false and the means are not all equal, then this measure of variance will be larger. The numerator is computed by measuring the variance of the means and if the true means of the groups are identical then this is a function of the overall variance of the data. If the null hypothesis is true then these are both estimates of the same thing and the ratio will be around 1. So these are the two sat scores, such that the middle 75% is contained between them.The F statistic is a ratio of 2 different measure of variance for the data. So this value was simply going to be at the positive version of this number. We know that the normal distribution is symmetric. Here corresponds with negative 2.24 and the one on the right corresponds with negative 2.25 So this value is somewhere between negative two point 24 and negative 2.25 So a good guess is going to be negative 2.245 for the value on the left and then this value on the right. These are the closest to 0.125 on this table on this one on the left. So we find that our answer is going to be somewhere between these two values. So we go back to our normal table, you scroll up and we're looking for 0.1 to 5. There corresponds with an area of 0.1 to 5. So we're gonna find thes n score on our standard normal table. We can calculate the area in the tales, so we know that the total area needs to add upto one, which tells us that in either tail we're gonna have 12.5% or 0.1 to 5. It's going to be Sarah 0.75 in the middle and there's gonna be two zone scores that bound to this area. That's where 0.85 So we know where answer is gonna be somewhere between 0.84 and 0.85 So a good guess would be 0.845 And for B the middle 75%. That score of 0.84 and this one corresponds with us. So the one on the left here corresponds with us. So the closest values on here 20.8 are these two values right here. We see that these values are all way too small. So in order to find this, we're going to use our Sandra normal table and we're going to look for the area that corresponds to 0.8.

So in other words, the area to the left of songs that score is equal to 0.8. The important thing to know about the 80th percentile is that it is the value such but 80% of values are less than it.

How to find z score on minitab express