Answer to Question #341736 in Statistics and Probability for Velle

Question

On an enquiry related to the frequent reporting of mouth cancer cases, it is estimated that around 60% of the patients chewed tobacco. Does this seem to be a valid estimate if, in a random sample of 120 people, 84 were found to be tobacco users?

Answer 1

The following null and alternative hypotheses for the population proportion needs to be tested:

$H_0:p=0.6$

$H_a:p\not=0.6$

This corresponds to a two-tailed test, for which a z-test for one population proportion will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a two-tailed test is $z_c = 1.96.$

The rejection region for this two-tailed test is $R = \{z: |z| > 1.96\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}=\dfrac{\dfrac{84}{120}-0.6}{\sqrt{\dfrac{0.6(1-0.6)}{120}}}$$

$$=2.236$$

Since it is observed that $|z| = 2.236 > 1.96=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=2P(Z>2.236)=0.025352,$ and since $p=0.025352<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population proportion $p$

is different than 0.6, at the $\alpha = 0.05$ significance level.