In determining whether or not to reject the null hypothesis, one method you can use is to find rejection regions. A rejection region is an area or range of values
under a standard normal distribution, where the null hypothesis is not probable. As a pure visual of this, here are graphs for a left tailed, a right tailed and a 2 tailed test.
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 the critical values are often noted as z naught for a one tailed test, and negative z naught
and positive z naught for 2 tailed tests. So, lets say that you have a random sample with a sample size greater than 30 from a population. And from this sample
you calculate a sample mean, x bar. So, if you found a standardized test statistic, the green line, which is often noted as z, whose value fell in a shaded area,
you would reject the null hypothesis, and if the standardized test statistic, again the green line did not fall in a shaded area, you would fail to reject the null hypothesis.
Lets go through an example to show how this works. Lets say that you have a job offer in a city located across the country, and you are concerned about the cost
of living in that city. You read a report that the mean cost of living per month in that city is $2800. You believe this information is incorrect, you claim that the mean cost
of living is less than $2800. So, you go out and gather data about the cost of living and find that in a random sample of 35 the mean is $2690 for the
cost of living per month. Assume the standard deviation is $400, at a level of confidence, alpha, of 0.05, test your claim. Step 1 is to make sure 2 conditions are met.
First 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 our sample size is 35, 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, mu is less than 2800. And we know the alternative hypothesis contains a statement of inequality, so h sub a is mu
less than 2800. The null hypothesis is the complement of the alternative hypothesis and contains a statement of equality, so h sub 0 is mu greater than or equal to 2800.
Step 3 is to identify the level of significance, which was given, alpha equals 0.05. Step 4, is to determine the test to use, left tailed, right tailed or 2 tailed,
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 z naught,
in a z distribution table where the area to the left equals 0.05. And in the table that value is -1.645, so that is our critical value, z 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 z naught, which is any value that is less than -1.645. This is often written as z less than z naught, so z less than -1.645. Step 7 is to use the formula and calculate the
z value, or the value of the standardized test statistic. And the formula for this is z equals x bar minus mu, divided by sigma over the square root of n.
And in our example, we have x bar, the sample mean equals 2690, mu, the hypothesized population mean equals 2800, sigma, the population standard deviation
equals 400, and n, the sample size equals 35. So, we can plug these into the formula and we get z equals -1.627. Step 8 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 not fall in the rejection region, as z, the standardized test statistic
is greater than z 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 support the claim that the mean cost of living in the city is less than $2800 per month.
One more example. A researcher claims the mean amount of money spent on food per household is $550 per month. In a random sample of 80 households the
mean amount of money spent on food per household was $588 per month. Assume a standard deviation of 130, at a level of confidence of alpha equals 0.10, test the claim.
Step 1, it is a random sample, and n, the sample size is greater than or equal to 30, so we are good there. Step 2, the claim is that mu equals 550.
We know that the null hypothesis contains the statement of equality, so h sub 0 is mu equals 550. And the alternative hypothesis is the complement of the null hypothesis
and contains a statement of inequality, so h sub a is mu does not equal 550. Step 3, the level of significance, was given, as alpha equals 0.10.
Step 4, to determine the test to use, and because the alternative hypothesis contains the does not equals inequality, this will be a 2 tailed test. Step 5, 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 z naught
and positive z 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 z distribution table that value is -1.645 and positive 1.645,
so those are our critical values. Step 6 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 z naught, or any value that is greater than positive z naught, which is any value less than -1.645
or any value greater than positive 1.645. Step 7, use the formula and calculate the z value, or the value of the standardized test statistic. And in our example, we have x bar,
the sample mean equals 588, mu, the hypothesized population mean equals 550, sigma, the population standard deviation equals 130, and n, the sample size equals 80.
So, we can plug these into the formula and we get z equals 2.614. Step 8 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 z, the standardized test statistic is greater than positive z naught,
the critical value. So, in this case we would reject the null hypothesis. Step 9, to interpret the decision. There is not enough evidence at the 10% level of significance
to support the claim that the mean amount of money spent on food per household is $550 per month. Alright my friends, that be the basics on
how to use rejection regions and critical values when you do hypothesis testing and you know the standard deviation.
Hopefully this video helped you out, I do have more videos right there for you, till next time, I am outta here.