Respuesta :
Answer:
The margin of error for a 95% confidence interval of the mean cost of the first year of owning and caring for a cat is of $43.12.
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].
That is z with a pvalue of [tex]1 - 0.025 = 0.975[/tex], so Z = 1.96.
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 and n is the size of the sample.
Sample of 81
This means that [tex]n = 81[/tex]
What is the margin of error for a 95% confidence interval of the mean cost of the first year of owning and caring for a cat?
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]M = 1.96\frac{198}{\sqrt{81}}[/tex]
[tex]M = 43.12[/tex]
The margin of error for a 95% confidence interval of the mean cost of the first year of owning and caring for a cat is of $43.12.
