Mastering Mathematics Smartly by Wee Wen Shih

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Hypothesis testing

This document highlights different cases on hypothesis testing involving unknowns and it also provides well-illustrated examples.

Level of significance

Recently, a doubt was raised about a past-year question on the level of significance. Let me reproduce the question below, which appeared in sgforums on 21 Jan '09:

This was my response, which I'd like to share with all keen learners:

An additional commentary to the response:

Discussion of different cases based on an example

We will consider a single example and modify its conditions to discuss the different cases that are likely to be asked in any examination.

Take, for example, a question that leads us to the following pieces of information:
\text{H}_0: \mu = 55
\text{H}_1: \mu > 55
Level of significance: 10%
\overline x = 67.6, s^2 = 1072.9714 have been calculated from \sum x and \sum x^2.
n = 15 from a Normal population whose \sigma^2 is unknown.

Case 1: Carry out a test (Z or t) directly.

We may use the p-value method or the critical region method to determine whether we reject \text{H}_0 or otherwise.

Using the p-value method:
We reject \text{H}_0 if p-value \leq 0.10.
From GC using t-test, p-value = 0.0792, so we reject \text{H}_0.

Using the critical range method:
We reject \text{H}_0 if calculated t-value \geq 1.345.
The value of 1.345 is obtained by referring to the t-distribution table on MF15. The value of 1.345 corresponds to the row where \nu = 14 (14 degrees of freedom, because n = 15) and to the column where p = 0.90 (because level of significance = 10%).
From GC using t-test, calculated t-value = 1.49, so we reject \text{H}_0.

Remarks: 
(i) If we have been given a large n from any population whose \sigma^2 is unknown, the Z-test will be used instead, due to the application of the Central Limit Theorem.

(ii) If the question gives \sum (x - \overline x)^2, then we would compute the unbiased estimate of the population variance as s^2 = \frac{\sum (x - \overline x)^2}{n-1}.

(iii) If the question gives \sum (x - a) and \sum (x - a)^2, then we would compute the unbiased estimate of the population variance as s^2 = \frac{1}{n-1}\left[\sum(x-a)^2 - \frac{\{\sum(x-a)\}^2}{n}}\right].

Case 2: Level of significance is unknown.

Modification to the example: Level of significance is \alpha.
The student is asked to find the smallest level of significance at which the test would result in rejection of the null hypothesis.

Since we reject \text{H}_0 if p-value \leq \alpha and GC tells us that p-value = 0.0792, we take the smallest \alpha to be 7.92.

Case 3: Value of the population mean is unknown.

Modification to the example: \text{H}_0 = \mu_0 and \text{H}_1 > \mu_0.
The student is asked to find the set of possible values of \mu_0, given that the null hypothesis is rejected in favour of the alternative hypothesis.

Using the critical range method, we reject \text{H}_0 if calculated t-value \geq 1.345.
So \frac{\overline x - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{67.6 - \mu_0}{\sqrt{\frac{1072.9714}{15}}} > 1.345
\Rightarrow \mu_0 < 56.2.


 

Conclusions from Z-test and t-test

Assume that the null hypothesis is \mu = \mu_0 and that the alternative hypothesis is \mu \geq \mu_0.

Suppose the conclusion from the t-test is to reject the null hypothesis at \alpha% level of significance, are we going to reject the null hypothesis using the Z-test? Cambridge has asked this question in the past.

Now let us assume that the t-test does not reject the null hypothesis, what is the conclusion we obtain from using the Z-test?

The approach to these analytical questions is to consider the relative positions of t_{\alpha} (i.e. the value that begins the critical region in the t-test) and z_{\alpha} (i.e. the value that begins the critical region in the Z-test).

Details of discussion coming soon...