Answer to Question #298157 in Statistics and Probability for Usha

Question

Random sample of 400 men and 200 women is locality were whether they would like to have bus stop near their residence 200 men and 40 women are favor of proposal the test is there any significance diff b/w men and women at 5%.

Answer 1

Sample Proportion 1 $\hat{p_1}=200/400=0.5$

Favorable Cases 1 $X_1=200$

Sample Size 1 $n_1=400$

Sample Proportion 2 $\hat{p_2}=\dfrac{40}{200}=0.2$

Favorable Cases 2 $X_2=40$

Sample Size 2 $n_2=200$

The value of the pooled proportion is computed as

$$\bar{p}=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{200+40}{400+200}=0.4$$

Significance Level $\alpha=0.05$

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

$H_0:p_1=p_2$

$H_1: p_1\not=p_2$

This corresponds to a two-tailed test, and a z-test for two population proportions 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:

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$$z=\dfrac{\hat{p}_1-\hat{p}_2}{\sqrt{\bar{p}(1-\bar{p})(1/n_1+1/n_2)}}$$ $$\approx\dfrac{0.5-0.2}{\sqrt{0.4(1-0.4)(1/400+1/200)}}\approx7.071$$

Since it is observed that $|z|=7.071>1.96=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=2P(Z>7.071)\approx0,$ and since $p=0<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

Hence we conclude that there is difference between the men and women in their attitude towards the bus stop near their residence at the $\alpha=0.05$ significance level.