Respuesta :
Answer:
a) 128.75
b) The 99% confidence interval for the population mean amount spent annually on a debit card is between $7,661.25 and $7,918.75
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{1-0.99}{2} = 0.005[/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]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/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 and n is the size of the sample.
[tex]M = 2.575*\frac{500}{\sqrt{100}} = 128.75[/tex]
So the answer for a) is 128.75
The lower end of the interval is the sample mean subtracted by M. So it is 7790 - 128.75 = $7,661.25
The upper end of the interval is the sample mean added to M. So it is 7790 + 128.75 = $7,918.75
The 99% confidence interval for the population mean amount spent annually on a debit card is between $7,661.25 and $7,918.75
