Respuesta :
Answer:
The 99% confidence interval for the mean number of toys purchased each year is between 7.6 toys and 7.8 toys.
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 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{1.8}{\sqrt{7.7}} = 0.1[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 7.7 - 0.1 = 7.6 toys
The upper end of the interval is the sample mean added to M. So it is 7.7 + 0.1 = 7.8 toys
The 99% confidence interval for the mean number of toys purchased each year is between 7.6 toys and 7.8 toys.
Answer:
99% confidence interval for the mean number of toys purchased each year is [7.6 , 7.8].
Step-by-step explanation:
We are given that a toy manufacturer wants to know how many new toys children buy each year. A sample of 1417 children was taken to study their purchasing habits.
Also, the population standard deviation is 1.8.
So, the pivotal quantity for 99% confidence interval for the average age is given by;
P.Q. = [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\bar X[/tex] = sample mean = 7.7
[tex]\sigma[/tex] = population standard deviation = 1.8
n = sample of children = 1417
[tex]\mu[/tex] = population mean
So, 99% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-2.5758 < N(0,1) < 2.5758) = 0.99
P(-2.5758 < [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < 2.5758) = 0.99
P( [tex]-2.5758 \times {\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]{\bar X - \mu}[/tex] < [tex]2.5758 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ) = 0.99
P( [tex]\bar X -2.5758 \times {\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X +2.5758 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ) = 0.99
99% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X -2.5758 \times {\frac{\sigma}{\sqrt{n} } }[/tex] , [tex]\bar X +2.5758 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ]
= [ [tex]7.7 -2.5758 \times {\frac{1.8}{\sqrt{1417} } }[/tex] , [tex]7.7 +2.5758 \times {\frac{1.8}{\sqrt{1417} } }[/tex] ]
= [7.6 , 7.8]
Therefore, 99% confidence interval for the mean number of toys purchased each year is [7.6 , 7.8].
