Respuesta :
Answer:
The margin of error for a 95% confidence interval for the population mean is of 1.05 years.
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.95}{2} = 0.025[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]
Now, find the margin of error M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population(square root of the variance) and n is the size of the sample.
What is the margin of error for a 95% confidence interval for the population mean
36 students, so [tex]n = 36[/tex]
Variance of 10.24, so [tex]\sigma = \sqrt{10.24} = 3.2[/tex]
[tex]M = 1.96*\frac{3.2}{\sqrt{36}} = 1.05[/tex]
The margin of error for a 95% confidence interval for the population mean is of 1.05 years.
