Question #346591

A local political leader claims that 95% of the families in his area of responsibility 

were given “ayuda” during a one-month lockdown. Of a random sample of 200 families, 187 said they received relief or “ayuda”. Is this enough to affirm the 

leader’s claim? Use 𝑎 = 0.05.


1
Expert's answer
2022-06-01T11:13:44-0400

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

H0:p=0.95H_0:p=0.95

Ha:p0.95H_a:p\not=0.95

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

Evidence:

Based on the information provided, the significance level is α=0.05,\alpha = 0.05 , and the critical value for a two-tailed test is zc=1.96.z_c = 1.96.

The rejection region for this two-tailed test is R={z:z>1.96}.R = \{z: |z|> 1.96\}.

The z-statistic is computed as follows:


z=p^p0p0(1p0)n=187/2000.950.95(10.95)200=0.9733z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{187/200-0.95}{\sqrt{\dfrac{0.95(1-0.95)}{200}}}=-0.9733

Since it is observed that z=0.9733<1.96=zc,|z| =0.9733<1.96= z_c, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is p=2P(Z<0.9733)=0.330404,p=2P(Z<-0.9733)= 0.330404, and since p=0.330404>0.05=α,p=0.330404>0.05=\alpha, it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion pp is different than 0.95, at the α=0.05\alpha = 0.05 significance level.


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