Question

Assume out of 500 people 280 are rice eater and rest are wheat eater. Can
we say the proportion of rice and wheat eater are same at 95%?

Answer 1

For rice, we have that the sample size is $n_1=500,$ the number of favorable cases is $x_1=280,$ so then the sample proportion is $\hat{p}_1=\dfrac{280}{500}=0.56$

For wheat, we have that the sample size is $n_2=500,$ the number of favorable cases is $x_2=220,$ then the sample proportion is $\hat{p}_2=\dfrac{220}{500}=0.44$

The value of the pooled proportion is computed as

$$\bar{p}={x_1+x_2\over n_1+n_ 2}=0.5$$

Also, the given significance level is $\alpha=0.05$

The following null and alternative hypotheses need to be tested:

$H_0:p_1=p_2$

$H_1:p_1\not=p_2$

This corresponds to a two-tailed test, for which a z-test for two population proportions needs to be conducted.

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:

$$z={\hat{p}_1-\hat{p}_2 \over \sqrt{\bar{p}(1-\bar{p})(1/n_1+1/n_2)}}=$$

$$={0.56-0.44 \over \sqrt{0.5(1-0.5)(1/500+1/500)}}=3.795$$

Since it is observed that $|z|=3.795>1.96=z_c,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that population proportion $p_1$ is different than $p_2$​, at the 0.05 significance level.

Using the P-value approach: The p-value is $p=0.000148,$ and since $p=0.000148<0.05,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that population proportion $p_1$ is different than $p_2$, at the 0.05 significance level.