Asked by Kristen Khachikian on May 05, 2024
Verified
For a variable with 
= 15.00, how large should the sample be for a desired 95% confidence interval width of 6?
A) 10
B) 24
C) 96
D) 576
Sample Size
The number of observations or subjects used in a study or research, influencing the study's findings' reliability and validity.
Variable
An element, feature, or factor that is liable to vary or change, used in experiments to determine its effects on the outcome.
- Grasp the effects of sample size on the precision of confidence intervals.
Verified Answer
JM
Janette MarteMay 07, 2024
Final Answer :
C
Explanation :
In order to calculate the required sample size, we need to use the formula for margin of error:
Margin of error = z*(sigma/sqrt(n))
Where z is the z-score corresponding to the desired level of confidence, sigma is the population standard deviation (which we don't know, so we'll use a conservative estimate of 10), and n is the sample size.
We want the margin of error to be no more than 6, so we can set up the following inequality:
6 = z*(10/sqrt(n))
sqrt(n) = z*(10/6)
n = (z*(10/6))^2
Since we want a 95% confidence interval, the z-score is 1.96. Plugging this in, we get:
n = (1.96*(10/6))^2
n = 96.04
Rounding up, the required sample size is 97, which is closest to choice C: 96.
Margin of error = z*(sigma/sqrt(n))
Where z is the z-score corresponding to the desired level of confidence, sigma is the population standard deviation (which we don't know, so we'll use a conservative estimate of 10), and n is the sample size.
We want the margin of error to be no more than 6, so we can set up the following inequality:
6 = z*(10/sqrt(n))
sqrt(n) = z*(10/6)
n = (z*(10/6))^2
Since we want a 95% confidence interval, the z-score is 1.96. Plugging this in, we get:
n = (1.96*(10/6))^2
n = 96.04
Rounding up, the required sample size is 97, which is closest to choice C: 96.
Learning Objectives
- Grasp the effects of sample size on the precision of confidence intervals.