Alice works at a health care facility. She has measured the hemoglobin levels of 200 people. The data follows a normal distribut
1 answer:
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To evaluate a function at a given input, you have to substitute every occurrence of the variable with that particular value.
So, for the first function, you have
The name of the variable is obviously irrelelvant, so the same goes for the second function:
Answer:
The 90% confidence interval for the population proportion of all such firms with this as the primary motivation is (69.96%, 80.04%).
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 z-score that has a p-value of .
The Corporate Lawyer, a magazine for corporate lawyers, reported that out of 200 firms with employee stock ownership plans, 150 indicated that the primary reason for setting up the plan was tax related.
This means that
90% confidence level
So , z is the value of Z that has a p-value of
, so
.
The lower limit of this interval is:
The upper limit of this interval is:
As percent:
0.6996*100% = 69.96%
0.8004*100% = 80.04%.
The 90% confidence interval for the population proportion of all such firms with this as the primary motivation is (69.96%, 80.04%).