Respuesta :
Answer:
The 99% confidence interval for the mean number of computer games purchased each year is between 7.3 and 7.5 games.
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{1.4}{\sqrt{1233}} = 0.1[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 7.4 - 0.1 = 7.3.
The upper end of the interval is the sample mean added to M. So it is 7.4 + 0.1 = 7.5
The 99% confidence interval for the mean number of computer games purchased each year is between 7.3 and 7.5 games.
Answer: = ( 7.3, 7.5)
Therefore at 99% confidence interval (a,b) = ( 7.3, 7.5)
Step-by-step explanation:
Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.
The confidence interval of a statistical data can be written as.
x+/-zr/√n
Given that;
Mean gain x = 7.4
Standard deviation r = 1.4
Number of samples n = 1233
Confidence interval = 99%
z(at 99% confidence) = 2.58
Substituting the values we have;
7.4+/-2.58(1.4/√1233)
7.4+/-2.58(0.03987)
7.4+/-0.1028
7.4+/-0.1
= ( 7.3, 7.5)
Therefore at 99% confidence interval (a,b) = ( 7.3, 7.5)
