In this video we discuss how to do hypothesis tests using a t test (standard deviation unknown) using rejection regions and critical values. We go through the steps of the process using a couple of examples.
Transcript/notes (partial)
In a past video we discussed rejection regions and critical values as a method to determine whether or not to reject the null hypothesis.
Real quick review. Here are graphs for a left tailed, right tailed and 2 tailed tests. And these shaded areas would be rejection regions, and these red lines are critical values. The shaded areas are the levels of significance, noted as alpha in one tailed tests, and one half alpha for 2 tailed tests, and when the population standard deviation is unknown, the critical values are often noted as t naught for a one tailed test, and negative t naught and positive t naught for 2 tailed tests.
If a calculated standardized test statistic falls into a rejection region it would mean to reject the null hypothesis and a standardized test statistic that does not fall into a rejection region would mean to fail to reject the null hypothesis.
In a recent video we discussed how to do this when the population standard deviation is known, but, in many situations the population standard deviation, sigma, is unknown. In these situations you can use a t distribution with n minus 1 degrees of freedom to test the population mean, mew.
If you recall, the t distribution is a family of curves, which are determined by the degrees of freedom and the degrees of freedom are equal to n minus 1.
When we don’t know the population standard deviation and we want to conduct a hypothesis test for the population mean we use the formula t equals x bar minus mew divided by the quantity s over the square root of n.
In this formula, t is the standardized test statistic, x bar is the mean from a sample, mew is the population mean, which is being hypothesized, s is the sample standard deviation and n is the size of the sample.
A friend says that the mean price for the camera model you want is at least $1900. You think this is incorrect, and in a random sample of 19 cameras of that model, the mean price is $1785, and the standard deviation is $296. Assume the population is normally distributed, at a level of significance of alpha = 0.05, is there enough evidence to reject your friend’s claim?
Step 1 is to make sure the 2 conditions are met to use the t distribution. The sample must be a random sample, and it is as that was stated in the information given, and second, the population must be normally distributed or n, the sample size must be greater than or equal to 30, and it was stated that the population is normally distributed, so that condition is met.
Step 2 is to write the claim out and identify the null and alternative hypotheses. The claim is that the mean, greater than or equal to 1900. And so h sub 0 is mew is greater than or equal to 1900. So h sub a is mew is less than 1900.
Step 3 is to identify the level of significance, alpha = 0.05.
Step 4, is to determine the test to use, and because the alternative hypothesis contains the less than inequality, this will be a left tailed test.
Step 5 is to determine the critical value or values and since this is a one tailed test, a left tailed test there will be only 1 critical value. Since the level of significance, alpha equals 0.05, we need to find the value for t naught, in a t distribution table with n minus 1 degrees of freedom, which in this case is 19 - 1, so 18 degrees of freedom, where the area to the left equals 0.05. And in the table that value is -1.734, so that is our critical value, t naught.
Step 6 is to identify the rejection region, and our rejection region is any standardized test statistic value that falls in the shaded area, that is any value that is less than t naught, which is any value less than -1.734.
Step 7 is to use the formula and calculate the t value. So, we can plug these into the formula and we get t equals -1.693.
Step 8, On our graph, you can see that the standardized test statistic does not fall in the rejection region, as t, the standardized test statistic is greater than t naught, the critical value. So, in this case we would fail to reject the null hypothesis.
Step 9 is to interpret the decision. There is not enough evidence at the 5% level of significance to reject the claim that the mean cost of a used camera is at least $1900.
Timestamps
0:00 Review Of Rejection Regions And Critical Values
0:55 Use T Test When The Standard Deviation Is Unknown
1:17 T Test Formula For A Hypothesis Test
1:52 Example Problem 1 For Using A T Test For A Hypothesis Test
3:25 How To Find Critical Values For A T Test For Example 1
4:16 How To Calculate The T Value For Example 1
5:13 Example Problem 2 For Using A T Test For A Hypothesis Test
6:33 How To Find Critical Values For A T Test For Example 2
7:35 How To Calculate The T Value For Example 2