Remember that the t distribution is more wide than the standard normal distribution, but at n gets larger and larger, the t distribution become approximately normal in nature.

The t-test is used when population standard deviation is not known. When we know that population is normally distributed, then we can use Z and also when population standard deviation is known, we can use Z

Note that if you want to figure out if a function has a slant asymptote, you have to realize that the equation of a slant asymptote is linear, so the exponent of the leading coefficient of the numerator of the function must be one greater than that of the denominator.

When doing long division (denominator into numerator), it might not divide evenly. Do not concern yourself with the remainder. That slant or oblique asymptote is just the linear portion.

Suppose you roll two 6 sides dice and we want to see the outcomes and probability distribution for the difference between the two dice. 3 on the first and 2 on the second would be 3-2=1 and 2 on the first and 3 on the second would be 2-3 = -1

To get a 0 when subtracting the numbers on the dice, the numbers must be the same. This happens 6 ways.

(1,1), (2,2), (3,3), (4,4), (5,5), (6,6).. since there are 36 possibilities, we have 6/36 = 1/6

For problem 6, to subtract numbers to get -2, the second number must be two larger than the first number.

(1,3), (2,4), (3,5), (4,6) . So we have 4 out of 36 outcomes that will subtract to -2. The probability is 4/36 = 1/9

For problem 7, It's basically the same as problem 6, except now the first number is two larger than the second number.

This gives us (3,1), (4,2), (5,3), (6,4). Again this is 4/36, simplified to 1/9

For the last problem, we want the possible outcomes with corresponding probabilities

We can get 0 if the numbers are the same, which I showed in problem 5, that is probability 1/6

We can get 1 when subtracting the numbers if we have (2,1), (3,2), (4,3), (5,4), (6,5) that is 5/36

We can get -1 when subtracting if we have (1,2), (2,3), (3,4), (4,5), (5,6), again with probability 5/36

We can get 2 as shown in problem 7 with probability of 1/9
We can get -2 as shown in problem 6 with probability of 1/9
We can get 3 with rolls of (4,1), (5,2), (6,3) with probability of 3/36 = 1/12
We can get -3 with rolls (1,4), (2,5), (3,6) with probability of 3/36 = 1/12
We can get 4 with rolls (5,1), (6,2) with probability of 2/36 = 1/18
We can get -4 with rolls (1,5), (2,6) with probability of 2/36 = 1/18
We can get 5 with a roll of (6,1) with probability 1/36
We can get -5 with a roll of (1,6) with probability of 1/36

Putting it all together we get this model

outcome Probability
5 1/36
4 1/18
3 1/12
2 1/9
1 5/36
0 1/6
-1 5/36
-2 1/9
-3 1/12
-4 1/18
-5 1/36

Notice that the probabilities will add to 1. That must always be the case for a legitimate probability model and the probabilities for each event must be between 0 and 1 inclusive

Suppose you have the following distribution

x = 0, 2, 5

p(x) = 1/4, 1/4, 1/2

find the mean and variance.

To get the mean you take the sum of x(P(x))

so for x = 0, 1,5 with P(x) = 1/4, 1/4, and 1/2

You get 0(1/4) + 1(1/4) + 5(1/2) = 0 + 1/4 + 5/2 = 2.75

To calculate sigma squared (variance)

It's the [sum (x- mean)^2P(x)]/n

So we have (0 - 2.75)^2 + (2-2.75)^2 + (5-2.75)^2

The equals 7.5625 + 0.5625 + 5.0625 = 13.1875
now take 13.1875/3 = 4.396

For a sample size of two you can have these possibilities (0,0), (0,2), (0,5), (2,0), (2,2), (2,5), (5,0), (5,2), (5,5)

The means are the two numbers added and divided by two. That gives us
(0 + 0)/2 = 0
(0 + 2)/2 = 1
(0 + 5)/2 = 2.5
(2 + 0)/2 = 1
(2 + 2)/2 = 2
(2 + 5)/2 = 3.5
(5 + 0)/2 = 2.5
(5 +2)/2 = 3.5
(5 + 5)/2 + 5

So you can have mean of 0 with (0, 0) with probability (1/4)(1/4) = 1/16
mean of 1 with (0, 2) and (2,0) with probability 2(1/4)(1/4) = 1/8
mean of 2 with (2, 2) with probability of (1/4)(1/4) = 1/16
mean of 2.5 with (0,5) and (5,0) with probability of 2(1/4)(1/2) = 1/4
mean of 3.5 with (2,5) and (5,2) with probability of 2(1/4)(1/2) = 1/4
mean of 5 with (5,5) with probability of (1/2)(1/2) = 1/4

When conducting an ANOVA, it's important to realize that the test in and of itself will not tell which of the means in the null hypothesis are different from each other. The only thing we will know is whether or not all the means are equal or if one or more of the means are different. If you want to know which mean or means are different from the others you must perform one of the post hoc tests. You can find more information about these with a Google search on the topic.

Suppose you need to take a log transformation on set of data that is non-linear in nature.  You can tell this from a scatterplot.  Take ln (natural logarithm) of each x value and y value to complete the transformation. In doing so, you will get some errors in x for ln(0) and negative values for ln of values between 0 and 1. So to compensate for that add 1 to each x value before taking natural log. Then you will eliminate such problems and can get the regression equation and scatterplot accordingly

If we are now testing to see if two variances are equal, as compared to where we wanted to see if the variance equaled a certain value, we use an F test instead of a Chi-square test.  So the test statistic is an F statistic  equal to s1^2/s2^2 and we compare to an F critical value found in any F chart. The decision rule is reject if F falls in the rejection region and do not reject if F is not in the rejection region.  The hypotheses for this test would be Ho: variance1 = variance2,  Ha: variance1 does not equal variance2