Respuesta :
Answer:
the answer is option c
Step-by-step explanation:
i took the test
Testing the hypothesis, it is found that since the p-value of the test is of 0.0951 > 0.05, there is not significant evidence to conclude that the percentage is lower than 75%.
At the null hypothesis, it is tested if the proportion is of 75%, that is:
[tex]H_0: p = 0.75[/tex]
At the alternative hypothesis, it is tested if the proportion is of less than 75%, that is:
[tex]H_1: p < 0.75[/tex]
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.
For this problem, the parameters are: [tex]p = 0.75, n = 200, \overline{p} = \frac{142}{200} = 0.71[/tex].
The value of the test statistic is:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
[tex]z = \frac{0.71 - 0.75}{\sqrt{\frac{0.75(0.25)}{200}}}[/tex]
[tex]z = -1.31[/tex]
The p-value of the test is the probability of finding a sample proportion below 0.71, which is the p-value of z = -1.31.
Looking at the z-table, z = -1.31 has a p-value of 0.0951.
Since the p-value of the test is of 0.0951 > 0.05, there is not significant evidence to conclude that the percentage is lower than 75%.
A similar problem is given at https://brainly.com/question/24166849
