Respuesta :
Using the sample data to test the hypothesis, we have that:
- The actual number of respondents who rated themselves as above average drivers is 906.
- The p-value of the test is 0, which is less than the standard significance level of 0.05, thus we can conclude that the proportion respondents who rated themselves as above average drivers is greater than 0.7.
77% of the sample of 1176 adults rated themselves as above average drivers, that is:
[tex]0.77(1176) = 906[/tex].
Thus, the actual number of respondents who rated themselves as above average drivers is 906.
At the null hypothesis, we test if the proportion is [tex]\displaystyle \frac{7}{10} = 0.7[/tex], that is:
[tex]H_0: p = 0.7[/tex]
At the alternative hypothesis, we test if the proportion is greater than 70%, that is:
[tex]H_1: p \geq 0.7[/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.
- n is the size of the sample.
In this problem, the parameters are [tex]\overline{p} = 0.77, p = 0.7, n = 1176[/tex].
Thus, the value of the test statistic is:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
[tex]z = \frac{0.77 - 0.7}{\sqrt{\frac{0.7(0.3)}{1176}}}[/tex]
[tex]z = 5.24[/tex]
The p-value of the test is the probability of finding a sample proportion above 0.77, which is 1 subtracted by the p-value of z = 5.24.
Looking at the z-table, z = 5.24 has a p-value of 1.
1 - 1 = 0.
The p-value of the test is 0, which is less than the standard significance level of 0.05, thus we can conclude that the proportion respondents who rated themselves as above average drivers is greater than 0.7.
A similar problem is given at https://brainly.com/question/24330815
