For sample 1, we have that the sample proportion is "\\hat{p}_1=\\dfrac{X_1}{N_1}=\\dfrac{450}{600}=0.75."
For sample 1, we have that the sample proportion is "\\hat{p}_2=\\dfrac{X_2}{N_2}=\\dfrac{450}{900}=0.5."
The value of the pooled proportion is computed as
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:
"=\\dfrac{0.75-0.5}{\\sqrt{0.6(1-0.6)(\\dfrac{1}{600}+\\dfrac{1}{900})}}\\approx9.682"
Since it is observed that "|z|=9.682>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>9.682)=0," 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.
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