Answer to Question #121492 in Statistics and Probability for abidhusain

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.