In a past video we discussed how to test a claim comparing the means of 2 populations by using samples and comparing their means,
when we know the population standard deviations and the samples are independent, that is, the samples are not related. Real quick review, we use a z test in this case,
and the 4 conditions on the screen must be met to use the z test. We then mark critical values and rejection regions based on alpha, the level of significance
on a standard normal graph, and then use this formula to calculate a standardized test statistic from the samples, then mark that value on the graph,
and determine whether that value falls in a rejection region. In most cases, when testing a claim comparing the means of 2 populations, we do not know the population
standard deviations. The process for testing a claim in this situation is very similar to the process just mentioned. One difference is that condition 1 is changed
to the population standard deviations are unknown, the rest of the conditions remain the same, random samples, independent samples, and population normally distributed
or both sample sizes greater than or equal to 30. Another difference is that we are going to use a t test, so we use a t distribution table to find the critical values,
and the formula for calculating the standardized test statistic is t equals, x bar 1 minus x bar 2, minus mu 1, minus mu 2, divided by s xbar1 minus xbar2.
In this formula, t is the standardized test statistic, x bar 1 is the mean for sample 1, x bar 2 is the mean for sample 2, mu 1 is the mean for population 1,
mu 2 is the mean for population 2, and remember, we always assume mu1 minus mu 2 equals zero, so this part of the formula will always equal 0.
s x bar 1 minus x bar 2 has 2 possible formulas, depending on if the population variances, sigma squared 1 and sigma squared 2 are equal or not equal.
When the population variances are not equal s xbar1 minus xbar2 equals the square root of s1 squared, over n1, plus s2 squared, over n2 and the degrees of freedom
equal the smaller of n1 minus 1 or n2 minus 1. When the population variances are equal, s xbar1 minus xbar2 equals the square root of n1 minus 1, times s1 squared,
plus n2 minus 1, times s2 squared, divided by n1 plus n2, minus 2 ,times the square root of, 1 over n1 plus, 1 over n2, and the degrees of freedom equal n1
plus n2 minus 2. And in these formulas, s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the sample size of sample 1,
and n2 is the sample size of sample 2. Alright example time, and we will go through, step by step. 2 science teachers at rival schools have a friendly competition
with each other. Teacher 1 believes that her teaching methods are better and claims there is a difference in the mean science test scores for her students and her rival’s
students. Random samples for each of the teachers students test scores are listed in the table. Assume the populations are normally distributed and the population
variances are not equal. At a level of significance of alpha equals 0.10, test the claim. Step 1 is to make sure the 4 conditions are met. Both population standard deviations
are unknown, both samples are random samples as that was stated in the information given, the samples are independent, one sample from teacher 1 and one sample from
teacher 2, and the populations are normally distributed. So, all conditions are met. Step 2 is to write the claim out and identify the null and alternative hypotheses.
The claim is that there is a difference between the population means from teacher 1 and teacher 2, so mu 1 does not equal mu 2. And we know the alternative hypothesis
contains a statement of inequality, so h sub a is mu 1 does not equal mu 2. The null hypothesis is the complement of the alternative hypothesis and contains a statement
of equality, so h sub 0 is mu1 is equal to mu 2. Step 3 is to identify the level of significance, which was given, alpha equals 0.10. Step 4 is to identify the degrees of
freedom, which is the smaller of n1 minus 1 or n2 minus 1. And that results in 21 degrees of freedom. Step 5, is to determine the test to use, left tailed, right tailed or
2 tailed, and because the alternative hypothesis contains the does not equals inequality, this will be a 2 tailed test. Step 6 is to determine the critical value or values
and since this is a 2 tailed test, there will be 2 critical values. Graphically this looks like this, with our critical values, negative t naught and positive t naught, here and both of
these shaded areas in the tails being rejection regions. Since the level of significance is alpha equals 0.10, and this is a 2 tailed test, each of these rejection regions
will be one half of alpha, which is 0.05. And in the t distribution table that value is -1.721 and positive 1.721, so those are our critical values.
Step 7 is to identify the rejection regions, and our rejection regions is any standardized test statistic value that falls in the shaded area, that is any value that is less than
negative t naught, or any value that is greater than positive t naught, which is any value less than -1.721 or any value greater than positive 1.721.
Step 8 is to use the formula and calculate the t value, or the value of the standardized test statistic. And in our example, we have 245 minus 233 minus 0,
as we always assume mu 1 minus mu 2 equals zero, divided by the square root of 23.4 squared over 27, plus 18.9 squared over 22.
And this calculates out to t equals 1.986. Step 9 is to make a decision to reject or fail to reject the null hypothesis. On our graph, you can see that the standardized test
statistic does fall in the rejection region to the right, as t, the standardized test statistic is greater than positive t naught, the critical value.
So, in this case we would reject the null hypothesis. Step 10 is to interpret the decision. There is enough evidence at the 10% level of significance
to support the claim that the mean science test scores for teacher 1’s students and her rival’s students are different. As a quick note, using this same problem,
if the population variances were equal, we would use n1 plus n2 minus 2, and get 47 degrees of freedom, which would make the critical values -1.6779 and 1.6779.
And we would then use this formula to calculate the standardized test statistic, and plugging and calculating, the result is 1.943. So, as you can see, the critical values
do change as does the standardized test statistic. Alright my friends, that be the basics on hypothesis testing between means when you got no clue what the standard
deviation is. Hopefully this video did help you out, I do have more statistical videos right there for you, till next time, I am outta here.