Using the z-distribution, it is found that the data provides convincing evidence that the proportion has increased.
At the null hypotheses, it is tested if the proportion has not increased, that is, the difference of the 2017 proportion and the 2014 proportions is not positive, hence:
[tex]H_0: p_2 - p_1 \leq 0[/tex]
At the alternative hypothesis, it is tested if the proportion has increased, hence:
[tex]H_1: p_2 - p_1 > 0[/tex]
For each sample, they are given by:
[tex]p_1 = 0.197, s_1 = \sqrt{\frac{0.197(0.803)}{61}} = 0.0509[/tex]
[tex]p_2 = 0.383, s_2 = \sqrt{\frac{0.385(0.615)}{52}} = 0.0675[/tex]
Hence, for the distribution of differences, they are given by:
[tex]\overline{p} = p_2 - p_1 = 0.383 - 0.197 = 0.186[/tex]
[tex]s = \sqrt{s_1^2 + s_2^2} = \sqrt{0.0509^2 + 0.0675^2} = 0.0845[/tex]
It is given by:
[tex]z = \frac{\overline{p} - p}{s}[/tex]
In which p = 0 is the value tested at the null hypothesis.
Hence:
[tex]z = \frac{\overline{p} - p}{s}[/tex]
[tex]z = \frac{0.186 - 0}{0.0845}[/tex]
z = 2.2
Considering a right-tailed test, as we are testing if the proportion is higher than a value, the critical value at the 0.05 significance level is [tex]z^{\ast} = 1.645[/tex].
Since the test statistic is greater than the critical value for the left tailed test, the data provides convincing evidence that the proportion has increased.
More can be learned about the z-distribution at https://brainly.com/question/26454209