Respuesta :
Finding the p-value of the test using the z-distribution, as we are working with a proportion, it is found that this was a significant difference for levels of significance of 0.02 and lower.
What are the hypothesis test?
At the null hypothesis, we test if the proportion is of 50.3%, hence:
[tex]H_0: p = 0.503[/tex]
We have a two-tailed test, hence at the alternative hypothesis we test if the proportion is different, that is:
[tex]H_1: p \neq 0.503[/tex]
What is the test statistic?
The test statistic is given by:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
In which:
- [tex]\overline{p}[/tex] is the sample proportion.
- p is the proportion tested at the null hypothesis.
- n is the sample size.
In this problem, the parameters are:
[tex]n = 300, \overline{p} = \frac{171}{300} = 0.57, p = 0.503[/tex]
Hence:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
[tex]z = \frac{0.57 - 0.503}{\sqrt{\frac{0.503(0.497)}{300}}}[/tex]
[tex]z = 2.32[/tex]
What is the p-value?
Considering a two-tailed test, using a z-distribution calculator, the p-value is of 0.0203. Hence this was a significant difference for levels of significance of 0.02 and lower.
More can be learned about the z-distribution at https://brainly.com/question/26454209
