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Answer:
The confidence interval at the 90% confidence level for the true population proportion of southern California community homes meeting at least the minimum recommendations for earthquake preparedness is (0.3136, 0.3830).
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 interval [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].
For this problem, we have that:
511 homes were surveyed, of which 178 met the minimum recommendations for earthquake preparedness. This means that [tex]n = 511[/tex] and [tex]p = \frac{178}{511} = 0.3483[/tex].
Find the confidence interval at the 90% confidence level for the true population proportion of southern California community homes meeting at least the minimum recommendations for earthquake preparedness.
So [tex]\alpha = 0.10[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.10}{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}} = 0.3483 - 1.645\sqrt{\frac{0.3483*0.6517}{511}} = 0.3136[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3483 + 1.645\sqrt{\frac{0.3483*0.6517}{511}} = 0.3830[/tex]
The confidence interval at the 90% confidence level for the true population proportion of southern California community homes meeting at least the minimum recommendations for earthquake preparedness is (0.3136, 0.3830).
The confidence interval at the 90% confidence level for the true population proportion of southern California for the interval (0.3136, 0.3830)
What is a z-score?
The z-score is a numerical measurement used in statistics of the value's relationship to the mean of a group of values, measured in terms of standards from the mean.
Five hundred eleven (511) homes in a certain southern California community are randomly surveyed to determine if they meet minimal earthquake preparedness recommendations.
One hundred seventy-eight (178) of the homes surveyed met the minimum recommendations for earthquake preparedness and 333 did not.
P = 178/511 = 0.3483
The confidence interval at the 90% confidence level for the true population proportion of southern California community homes meets at least the minimum recommendations for earthquake preparedness.
n = 511
The value of z for the confidence level of 90%
[tex]z = 1 - \dfrac{\alpha }{2} = 1 - \dfrac{0.90}{2} = 0.55[/tex]
Then we have
[tex]p \pm z\sqrt{\dfrac{p(1-p)}{n}}[/tex]
The lower limit of this interval will be
[tex]0.3485 - 0.55\sqrt{\dfrac{0.3483(1-0.3483)}{511}} =0.3136[/tex]
The upper limit of this interval will be
[tex]0.3485 + 0.55\sqrt{\dfrac{0.3483(1-0.3483)}{511}} =0.3830[/tex]
More about the z-score link is given below.
https://brainly.com/question/13299273
