Hypothesis Testing The Difference Between Means Standard Deviations Unknown Independent Samples

In this video we cover how to do hypothesis testing the difference between means for independent samples when the population standard deviations are unknown. Transcript/notes (partial) The formula for calculating the standardized test statistic is t equals x bar 1 minus x bar 2 minus mew1 minus mew2, 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, mew 1 is the mean for population 1, mew 2 is the mean for population 2, and remember, we always assume mew 1 minus mew 2 equals zero, so this part of the formula will always = 0. S x bar 1 minus x bar 2 has 2 possible formulas, depending on if the population variances, sigma 1 and sigma 2 are equal or not equal. 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 = 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 mew 1 does not equal mew 2. And we know the alternative hypothesis contains a statement of inequality, so h sub a is mew 1 does not equal mew 2. The null hypothesis is the complement of the alternative hypothesis and contains a statement of equality, so h sub 0 is mew 1 is equal to mew 2. Step 3 is to identify the level of significance, which was given, alpha = 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, 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, use the formula and calculate the z value, or the value of the standardized test statistic. And in our example, we have 245 minus 233 minus 0, as we always assume mew 1 minus mew 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 z 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 there is a difference in the mean science test scores for teacher 1’s students and her rival’s students. Timestamps 0:00 Review Of Comparing Means From Samples 0:51 Conditions For Comparing Means When Standard Deviation Is Unknown 1:10 Formula For A T Test For Comparing Means 2:46 Example Problem For Comparing Means When Standard Deviation Is Unknown 4:28 How To Find Critical Values 5:03 How To Determine Rejection Regions 5:22 How To Calculate T Value For Comparing Means When Standard Deviation Is Unknown