Answer:
[tex]\bar X = 69.46-3.31= 66.15[/tex]
[tex]\bar X = 62.84+3.31= 66.15[/tex]
Because the confidence interval is defined as [tex](\bar X -ME, \bar X + ME)[/tex]
The best option is : D. 66.15
Step-by-step explanation:
Previous concepts
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
[tex]\bar X[/tex] represent the sample mean (variable of interest)
[tex]\mu[/tex] population mean
[tex]\sigma=6.50[/tex] represent the population standard deviation
n=100 represent the sample size
Assuming the X follows a normal distribution
[tex]X \sim N(\mu, \sigma=6.5)[/tex]
The sample mean [tex]\bar X[/tex] is distributed on this way:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
The confidence interval on this case is given by:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
We have a confidence interval given (62.84, 69.46). W ecan find the margin of error for this interval with this:
[tex]ME= \frac{69.46-62.84}{2}=3.31[/tex]
With the margin of error we can find the sample mean with this:
[tex]\bar X = 69.46-3.31= 66.15[/tex]
[tex]\bar X = 62.84+3.31= 66.15[/tex]
Because the confidence interval is defined as [tex](\bar X -ME, \bar X + ME)[/tex]
The best option is : D. 66.15
Answer: Option D.
Step-by-step explanation:
The given interval for population mean was 62.84 to 69.46.
The sample mean will be at the midpoint of the interval.
That is: [tex]\frac{\left(62.84+69.46\right)}{2}=66.15[/tex].
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