Test about proportions  (Page 2/3)

The research worker’s hypothesis $H_1:p<0.05$ is called the alternative hypothesis . Since $H_1:p<0.05$ does not completely specify the distribution, it is a composite hypothesis because it is composed of many simple hypotheses.

The rule of rejecting $H_0$ and accepting $H_1$ if $Y\le 5$ , and otherwise accepting $H_0$ is called a test of a statistical hypothesis .

It is clearly seen that two types of errors can be recorded

• Type I error: Rejecting $H_0$ and accepting $H_1$ , when $H_0$ is true;
• Type II error: Accepting $H_0$ when $H_1$ is true, that is, when $H_0$ is false.

Since, in the example above, we make a Type I error if $Y\le 5$ when in fact p =0.05. we can calculate the probability of this error, which we denote by $\alpha$ and call the significance level of the test . Under an assumption, it is $$\alpha =P\left(Y\le 5;p=0.05\right)={\displaystyle \sum _{y=0}^5\left(\begin{array}{l}200\\ y\end{array}\right){\left(0.05\right)}^y{\left(0.95\right)}^{200-y}}.$$ .

Since n is rather large and p is small, these binomial probabilities can be approximated extremely well by Poisson probabilities with $\lambda =200\left(0.05\right)=10.$ That is, from the Poisson table, the probability of the Type I error is $$\alpha \approx {\displaystyle \sum _{y=0}^5\frac{{10}^y{e}^{-10}}{y!}}=\mathrm{0.067.}$$

Thus, the approximate significance level of this test is $\alpha =0.067$ . This value is reasonably small. However, what about the probability of Type II error in case p has been improved to 0.02, say? This error occurs if $Y>5$ when, in fact, p =0.02; hence its probability, denoted by $\beta$ , is $$\beta =P\left(Y>5;p=0.02\right)={\displaystyle \sum _{y=6}^{200}\left(\begin{array}{l}200\\ y\end{array}\right){\left(0.02\right)}^y{\left(0.98\right)}^{200-y}}.$$

Again we use the Poisson approximation, here $\lambda \text{=200(0}\text{.02)=4}$ , to obtain $$\beta \approx 1-{\displaystyle \sum _{y=0}^5\frac{4^y{e}^{-4}}{y!}}=1-0.785=\mathrm{0.215.}$$

The engineers and the statisticians who created this new procedure probably are not too pleased with this answer. That is, they note that if their new procedure of manufacturing circuits has actually decreased the probability of failure to 0.02 from 0.05 (a big improvement), there is still a good chance, 0.215, that ${\text{H}}_{\text{0}}\text{: p=0}\text{.05 }$ is accepted and their improvement rejected. Thus, this test of ${\text{H}}_{\text{0}}\text{: p=0}\text{.05 }$ against ${\text{H}}_{\text{1}}\text{: p=0}\text{.02 }$ is unsatisfactory. Without worrying more about the probability of the Type II error, here, above was presented a frequently used procedure for testing ${\text{H}}_{\text{0}}{\text{: p=p}}_{\text{0}}$ , where ${\text{p}}_{\text{0}}$ is some specified probability of success. This test is based upon the fact that the number of successes, Y , in n independent Bernoulli trials is such that $\text{Y/n}$ has an approximate normal distribution, ${\text{N[p}}_{\text{0}}{\text{, p}}_{\text{0}}{\text{(1- p}}_{\text{0}}\text{)/n]}$ , provided ${\text{H}}_{\text{0}}{\text{: p=p}}_{\text{0}}$ is true and n is large. Suppose the alternative hypothesis is ${\text{H}}_{\text{0}}{\text{: p>p}}_{\text{0}}$ ; that is, it has been hypothesized by a research worker that something has been done to increase the probability of success. Consider the test of ${\text{H}}_{\text{0}}{\text{: p=p}}_{\text{0}}$ against ${\text{H}}_{\text{1}}{\text{: p> p}}_{\text{0}}$ that rejects ${\text{H}}_{\text{0}}$ and accepts ${\text{H}}_{\text{1}}$ if and only if

$$Z=\frac{Y/n-p_0}{\sqrt{p_0\left(1-p_0\right)/n}}\ge z_{\alpha }.$$

That is, if $\text{Y/n}$ exceeds ${\text{p}}_{\text{0}}$ by standard deviations of $\text{Y/n}$ , we reject ${\text{H}}_{\text{0}}$ and accept the hypothesis ${\text{H}}_{\text{1}}{\text{: p> p}}_{\text{0}}$ . Since, under ${\text{H}}_{\text{0}}$ Z is approximately $\text{N}\left(\text{0,1}\right)$ , the approximate probability of this occurring when ${\text{H}}_{\text{0}}{\text{: p=p}}_{\text{0}}$ is true is $\alpha$ . That is the significance level of that test is approximately $\alpha$ . If the alternative is ${\text{H}}_{\text{1}}{\text{: p< p}}_{\text{0}}$ instead of ${\text{H}}_{\text{1}}{\text{: p> p}}_{\text{0}}$ , then the appropriate $\alpha$ -level test is given by $Z\le -z_{\alpha }$ . That is, if $\text{Y/n}$ is smaller than ${\text{p}}_{\text{0}}$ by standard deviations of $\text{Y/n}$ , we accept ${\text{H}}_{\text{1}}{\text{: p< p}}_{\text{0}}$ .