Using the z-distribution, as we are working with a proportion, it is found that the p-value of the test is of 0.
At the null hypothesis, it is tested if the proportion is of 71%, that is:
[tex]H_0: p = 0.71[/tex]
At the alternative hypothesis, it is tested if the proportion has decreased, that is:
[tex]H_1: p < 0.71[/tex].
The test statistic is given by:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
In which:
In this problem, we have that the parameters are:
[tex]n = 300, \overline{p} = \frac{165}{300} = 0.55[/tex]
Hence, the value of the test statistic is found as follows:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
[tex]z = \frac{0.55 - 0.71}{\sqrt{\frac{0.71(0.29)}{300}}}[/tex]
z = -6.11
Using a z-distribution calculator, with a left-tailed test, as we are testing if the proportion is less than a value, and z = -6.11, it is found that the p-value is of 0.
More can be learned about the z-distribution at https://brainly.com/question/26454209
