Respuesta :
Answer:
The sample proportion represents a statistically significant difference from 50%
Step-by-step explanation:
Null hypothesis: The sample proportion is the same as 50%
Alternate hypothesis: The sample proportion is not the same as 50%
z = (p' - p) ÷ sqrt[p(1 - p) ÷ n]
p' is sample proportion = 289/400 = 0.7225
p is population proportion = 50% = 0.5
n is number of students sampled = 400
z = (0.7225 - 0.5) ÷ sqrt[0.5(1 - 0.5) ÷ 400] = 0.2225 ÷ 0.025 = 8.9
The test is a two-tailed test. Using a 0.01 significance level, critical value is 2.576. The region of no rejection of the null hypothesis is -2.576 and 2.576.
Conclusion:
Reject the null hypothesis because the test statistic 8.9 falls outside the region bounded by the critical values -2.576 and 2.576.
There is sufficient evidence to conclude that the sample proportion represents a statistically significant difference from 50%.
Testing the hypothesis, we can conclude that since the p-value of the test is 0 < 0.01, we can conclude that this represents a statistically significant difference from 50% using a significance level of 0.01.
At the null hypothesis, we test if the proportion is of 0.5, that is:
[tex]H_0: p = 0.5[/tex]
At the alternative hypothesis, we test if the proportion is different of 0.5, that is:
[tex]H_1: p \neq 0.5[/tex]
The test statistic is given by:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
In which:
- [tex]\overline{x}[/tex] is the sample proportion.
- p is the proportion tested at the null hypothesis.
- n is the sample size.
In this problem:
[tex]p = 0.5, n = 400, \overline{p} = \frac{289}{400} = 0.7225[/tex]
The value of the test statistic is:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
[tex]z = \frac{0.7225 - 0.5}{\sqrt{\frac{0.5(0.5)}{400}}}[/tex]
[tex]z = 8.9[/tex]
We have a two-tailed test, as we are testing if the proportion is different from a value, thus the p-value is P(|z| > 8.9), which is 2 multiplied by the p-value of z = -8.9.
z = -8.9 has a p-value of 0.
Since the p-value of the test is 0 < 0.01, we can conclude that this represents a statistically significant difference from 50% using a significance level of 0.01.
A similar problem is given at https://brainly.com/question/17100502
