In recent years, the mean age of all college students in city X has been 23. This year,
a random sample of 42 students revealed a mean age of 23.8. Suppose their ages are normally
distributed with a population standard deviation of 2.4. Can we infer at α = 0.05 that the
population mean age has changed? Since we’re testing if the population mean
age has changed from 23, the alternate hypothesis will be that the mean is not equal to 23.
And the null hypothesis mean equals to 23. The significance level α is 0.05 or 95% confidence.
Since this is a two-tailed test, the rejection region will be in both tails of the distribution.
So we divide alpha by 2 to have .025 in each tail.
Next we find the critical values. That is, the values that separate the rejection
region from the non-rejection region. For single population mean tests, we either
conduct a z -test or a t -test. If the population standard deviation σ is
known, a z-test is employed, and if the population standard deviation is unknown, a t -test is
used. In some courses, however, due to the central
limit theorem, whenever the sample size is large, and some textbooks say when n is at
least 30, apply the z test whether the population standard deviation is known or not.
The onus is on you to check what obtains in your course.
For this example, we will conduct a z-test because the population standard deviation
σ is given. We now find z critical values that define
the critical or rejection region using the standard normal table.
The 2 commonly used z tables are the cumulative less than tables (with negative and positive
z values), and the cumulative from the mean tables.
These pictures are usually displayed at the top of the respective tables.
To find the critical value on the cumulative less-than table, on the negative side, since
the area in the left tail is 0.025, we then search for the closest value to 0.025 in the
body of the table. We actually have 0.0250 here corresponding to a z-score of -1.96.
And because of symmetry, the positive z-score is also 1.96.
In the mean to z table, the entries represent the area from 0 to z, shaded here. And because
we have 0.025 in the tails, and the area on either side of the mean is 0.5, then the area
from 0 to z is 0.5 minus .025 which is 0.475. We then search for the closest to .475 in
the table. And we have the actual value here, corresponding to a z score of 1.96. And because
of symmetry, the z score on the left will be -1.96.
Sometimes we can also use the student t distribution table to find the z critical value.
When you scroll to the bottom of most t-tables, you either find infinity or z on the last
row here. This is because the t-distribution approaches the z distribution as the sample
size becomes very large. So these are actually z critical values.
And if you check under .025 in one tail or .05 in 2 tails, you will see that the corresponding
critical value here is 1.96. So back to the hypothesis test.
Based on the critical values found in the tables we can set a decision rule as follows:
Reject the null hypothesis if the test statistic z, based on this formula, is less than -1.96
or if it is greater than positive 1.96. So using the sample mean, population mean,
population standard deviation, and the sample size, we calculate the test statistic as 2.16.
And now its decision time. Since z = 2.16 is greater than 1.96, we reject
the null hypothesis. Because the null hypothesis which states that
the mean age is equal to the 23 has been rejected, we then have to support the alternative that
states that the mean age has changed. So we say we have enough evidence to conclude
that the mean age has changed, at α = 0.05. Now, suppose we conduct this same test at
2% significance level. The null and alternative hypotheses are still
the same. The significance level α is 0.02.
Dividing alpha into the two tails we have 0.01 in each tail.
Looking this up in the t table, under .01 one tail or .02 2-tails, we see that the critical
value is 2.326, which we can round to 2.33.
So we reject the null hypothesis if the test statistic is less than - 2.33 or if it is
greater than +2.33. The test statistic remains 2.16 which does
not fall into the rejection region. We therefore fail to reject the null hypothesis.
Consequently, we cannot support the alternative. Therefore, at α = 0.02, there is not enough
evidence to infer that the mean age has changed. And that concludes this tutorial.
Thanks for watching.