Solution to Random sample of 400 men and 200 women is locality were whether they would like … - Sikademy
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Archangel Macsika

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%.

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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


Significance Level \alpha=0.05

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


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:


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.

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Question ID: mtid-4-stid-46-sqid-2115-qpid-585