Respuesta :
Using the z-distribution, it is found that the 90% confidence interval for the percentage of successful challenges is (17.28, 29.34).
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which z is the z-score that has a p-value of [tex]\frac{1+\alpha}{2}[/tex].
31 out of 133 challenges were successful, hence:
[tex]n = 133, \pi = \frac{31}{133} = 0.2331[/tex]
90% confidence level, hence [tex]\alpha = 0.9[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.9}{2} = 0.95[/tex], so [tex]z = 1.645[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2331 - 1.645\sqrt{\frac{0.2331(0.7669)}{133}} = 0.1728[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2331 + 1.645\sqrt{\frac{0.2331(0.7669)}{133}} = 0.2934[/tex]
As percentages:
0.1728 x 100% = 17.28%
0.2934 x 100% = 29.34%
The 90% confidence interval for the percentage of successful challenges is (17.28, 29.34).
A similar problem is given at https://brainly.com/question/16807970
