Respuesta :
Answer:
The 90% confidence interval for the proportion of blocks that are sufficiently strong is (0.849, 1).
Step-by-step explanation:
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]1-\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 zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
Twenty six of the 28 blocks were sufficiently strong.
This means that [tex]n = 28, \pi = \frac{26}{28} = 0.929[/tex]
90% confidence level
So [tex]\alpha = 0.1[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.1}{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}} = 0.929 - 1.645\sqrt{\frac{0.929*0.071}{28}} = 0.849[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.929 + 1.645\sqrt{\frac{0.929*0.071}{28}} = 1.009[/tex]
Since a proportion cannot be above 1, the upper limit is 1.
The 90% confidence interval for the proportion of blocks that are sufficiently strong is (0.849, 1).
