Question #334779

In a certain Barangay, 550 random residents are chosen in which 225 are males. At a=0.01, test the hypothesis that more than 35% of the Barangay population are male residents.

1
Expert's answer
2022-04-29T05:41:43-0400

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

H0:p0.35H_0:p\le0.35

Ha:p>0.35H_a:p>0.35

This corresponds to a right-tailed test, for which a z-test for one population proportion will be used.

The z-statistic is computed as follows:


z=p^p0p0(1p0)n=2255500.350.35(10.353)5502.9054z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{\dfrac{225}{550}-0.35}{\sqrt{\dfrac{0.35(1-0.353)}{550}}}\approx2.9054

The p-value is p=P(Z>0.0970)=0.001834p =P(Z>0.0970)= 0.001834

Based on the information provided, the significance level is α=0.01,\alpha = 0.01, and the critical value for a right-tailed test is zc=2.3263.z_c = 2.3263.

The rejection region for this right-tailed test is R={z:z>2.3263}.R = \{z: z> 2.3263\}.

Since it is observed that z=2.90542.3263=zc,z = 2.9054 \ge2.3263= z_c , it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.922726,p =0.922726, and since p=0.0018340.01=α,p = 0.001834 \le 0.01=\alpha, it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population proportion pp is greater than 0.35, at the α=0.01\alpha = 0.01 significance level.



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