Answer to Question #346591 in Statistics and Probability for John Lloyd

Question

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.

Answer 1

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

$H_0:p=0.95$

$H_a:p\not=0.95$

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 $\alpha = 0.05 ,$ 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:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{187/200-0.95}{\sqrt{\dfrac{0.95(1-0.95)}{200}}}=-0.9733$$

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

Using the P-value approach:

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

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