Respuesta :
Answer:
[tex] ME=1.998 \frac{4}{\sqrt{64}}=0.999 \approx 1[/tex]
Step-by-step explanation:
Previous concepts
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".
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".
[tex]\bar X =69[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean
s=4.0 represent the sample standard deviation
n=64 represent the sample size
Assuming the X follows a normal distribution
[tex]X \sim N(\mu, \sigma)[/tex]
The sample mean [tex]\bar X[/tex] have the following distribution:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
The margin of error for the sample mean is given by this formula:
[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]
Since we have the sample standard deviation we can estimate the margin of error like this:
[tex] ME=t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]
The next step would be find the value of [tex]\z_{\alpha/2}[/tex], [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex]
We can find the degrees of freedom like this:
[tex]df=n-1=64-1=63[/tex]
And we can find the critical value with the following code in excel for example: "=T.INV(0.025,63)" and we got:
[tex]t_{\alpha/2}=\pm 1.998[/tex]
And the margin of error would be :
[tex] ME=1.998 \frac{4}{\sqrt{64}}=0.999 \approx 1[/tex]
