Question #341736

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?


1
Expert's answer
2022-05-17T09:59:21-0400

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

H0:p=0.6H_0:p=0.6

Ha:p0.6H_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 α=0.05,\alpha = 0.05, and the critical value for a two-tailed test is zc=1.96.z_c = 1.96.

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

The z-statistic is computed as follows:


z=p^pp(1p)n=841200.60.6(10.6)120z=\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=2.236

Since it is observed that z=2.236>1.96=zc,|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,p=2P(Z>2.236)=0.025352, and since p=0.025352<0.05=α,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 pp

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


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