The statement that 99% of all confidence intervals with a 99% confidence level should contain the population parameter of interest is false.
A confidence interval (CI) is essentially a range of estimates for an unknown parameter in frequentist statistics. The most frequent confidence level is 95%, but other levels, such 90% or 99%, are infrequently used for generating confidence intervals.
The confidence level is a measurement of the proportion of long-term associated CIs that include the parameter's true value. This is closely related to the moment-based estimate approach.
In a straightforward illustration, when the population mean is the quantity that needs to be estimated, the sample mean is a straightforward estimate. The population variance can also be calculated using the sample variance. Using the sample mean and the true mean's probability.
Hence we can generally infer that the given statement is false.
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Answer:
This is not a real question rephrase
Step-by-step explanation:
Answers:
x = 4
EF = 14
CF = 7
EC = 7
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Work Shown:
C is the midpoint of segment EF. This means that EC = CF. In other words, the two pieces are congruent.
Use substitution and solve for x
EC = CF
5x-13 = 3x-5
5x-13+13 = 3x-5+13
5x = 3x+8
5x-3x = 3x+8-3x
2x = 8
2x/2 = 8/2
x = 4
Now that we know that x = 4, we can use this to find EC and CF
Let's compute EC
EC = 5x - 13
EC = 5*x - 13
EC = 5*4 - 13 ... replace x with 4
EC = 20 - 13
EC = 7
Let's compute CF
CF = 3x - 5
CF = 3*x - 5
CF = 3*4 - 5 ... replace x with 4
CF = 12 - 5
CF = 7
As expected, EC = CF (both are 7 units long).
By the segment addition postulate, we can say EC+CF = EF
EC+CF = EF
EF = EC+CF
EF = 7+7
EF = 14
