Answer:
The 95% confidence interval for the population proportion is (0.1456, 0.2344). This means that we are 95% sure that the true proportion of employed American who say that they would fire their boss if they could is between 0.1456 and 0.2344.
Step-by-step explanation:
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:
95% 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:
The 95% confidence interval for the population proportion is (0.1456, 0.2344). This means that we are 95% sure that the true proportion of employed American who say that they would fire their boss if they could is between 0.1456 and 0.2344.
120138 subtract both and that would be the answer
The answer is B! Good luck!
The given sample's confidence interval is 4.73 ± 0.1199, or from 4.61 to 4.85
<h3>What is the z score?</h3>
The z-score is a numerical assessment of a value's connection to the mean of a set of values, expressed in terms of standards from the mean, that is used in statistics.
Given data;
Mean = 4.73
Standard deviation = 0.865
Sample size = 200
interval = 95%
Confidence interval=?
95% of samples contain the population mean (μ) within the confidence interval of 4.73 ± 0.1199.
Hence, the sample's confidence interval is 4.73 ± 0.1199, from 4.61 to 4.85
To learn more about the Z score, refer to;
brainly.com/question/15016913
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She has been gone 52 min.
57 - 5 = 52
