Respuesta :
Answer:
[tex] ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]
The lenght of the interval correspond to:
[tex] 8.4 = 2ME[/tex]
[tex] ME= \frac{8.4}{2}= 4.2[/tex]
And since we know the margin of error we can find the limits for the confidence interval:
[tex] Lower = 10 -4.2=5.8[/tex]
[tex] Upper = 10 +4.2=14.2[/tex]
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=10[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma[/tex] represent the population standard deviation
n represent the sample size
Solution to the problem
Assuming the X follows a normal distribution
[tex]X \sim N(\mu, \sigma)[/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)
The margin of error is given by:
[tex] ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]
The lenght of the interval correspond to:
[tex] 8.4 = 2ME[/tex]
[tex] ME= \frac{8.4}{2}= 4.2[/tex]
And since we know the margin of error we can find the limits for the confidence interval:
[tex] Lower = 10 -4.2=5.8[/tex]
[tex] Upper = 10 +4.2=14.2[/tex]
