I believe some of the brainly admins are abusing their power because they keep deleting my questions and I am not violating guid
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Answer:
The 99% confidence interval for the percentage of Phoenix residents who support mandatory sick leave for food handlers is between 85.40% and 92.60%.
Step-by-step explanation:
Confidence interval for the proportion:
In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of .
For this problem, we have that:
99% confidence level
So , z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:
The upper limit of this interval is:
For the percentage:
Multiply the proportion by 100.
0.8540*100 = 85.40%
0.9260*100 = 92.60%
The 99% confidence interval for the percentage of Phoenix residents who support mandatory sick leave for food handlers is between 85.40% and 92.60%.
The asymptotes are in the picture
