Respuesta :
Answer:
a
The sample size is [tex]n = 219.2[/tex]
b
The sample size is [tex]n = 537.5[/tex]
Step-by-step explanation:
From the question we are told that
The standard deviation is [tex]\sigma = 45 \minutes[/tex]
The Margin of Error is [tex]E = \pm 5 \ minutes[/tex]
Generally the margin of error is mathematically represented as
[tex]E = z * \frac{\sigma }{\sqrt{n} }[/tex]
Where n is the sample size
So
[tex]n = [\frac{z * \sigma }{E} ]^2[/tex]
Now at 90% confidence level the z value for the z-table is
z = 1.645
So
[tex]n = [\frac{1.645 * 45 }{5} ]^2[/tex]
[tex]n = 219.2[/tex]
The z-value at 99% confidence level is
[tex]z = 2.576[/tex]
This is obtained from the z-table
So the sample size is
[tex]n = [\frac{2.576 * 45 }{5} ]^2[/tex]
[tex]n = 537.5[/tex]
For the 90% confidence interval, the sample size is 219.2 and for the 99% confidence interval, the sample size is 537.5 and this can be determined by using the formula of margin of error.
Given :
- An advertising executive wants to estimate the mean amount of time that consumers spend with digital media daily.
- From past studies, the standard deviation is estimated as 45 minutes.
The formula of the margin of error can be used in order to determine the sample size is needed if the executive wants to be 90% confident of being correct to within 5 minutes.
[tex]\rm ME = z\times \dfrac{\sigma}{\sqrt{n} }[/tex]
For the 90% confidence interval, the value of z is 1.645.
Now, substitute the values of all the known terms in the above formula.
[tex]\rm n=\left(\dfrac{z\times \sigma}{ME}\right)^2[/tex] --- (1)
[tex]\rm n=\left(\dfrac{1.645\times 45}{5}\right)^2[/tex]
n = 219.2
Now, for 99% confidence interval, the value of z is 2.576.
Again, substitute the values of all the known terms in the expression (1).
[tex]\rm n=\left(\dfrac{2.576\times 45}{5}\right)^2[/tex]
n = 537.5
For more information, refer to the link given below:
https://brainly.com/question/6979326
