The size of the rat population of a wharf area grows at a rate of 5% monthly. If there are
1 answer:
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Answer:
The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).
The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).
Step-by-step explanation:
We have the standard deviations for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 85 - 1 = 84
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 84 degrees of freedom(y-axis) and a confidence level of . So we have T = 1.989.
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.26 = 4.24 days
The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.26 = 4.76 days
The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).
92% confidence interval:
Following the sample logic, the critical value is 1.772. So
The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.23 = 4.27 days
The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.23 = 4.73 days
The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).
Step-by-step explanation:
Erase the dot points you already have. We are supposed to substitute those values in the right side of problem 1 into the function as x.
For example if x=-4
If x=-2
If x=0
So our point should be
-4,9
-2,7
0,5
-2,3
-4,1.
The range is all possible y values in a function. Since this is discrete and we are given the domain, our range will just be the y value of the points you graphed.
(9,7,5,3,1)