Respuesta :
Answer:
a) The 95% confidence interval for the true mean is:
[tex]8.0695\leq \mu\leq9.2025[/tex]
b) This margin of error can be achieved with a sample of size 27 individuals.
Step-by-step explanation:
a) We have to construct a 95% confidence interval for the populations mean WBC.
The sample results, from a sample size n=187, are:
- Sample mean: 8.636
- Standard deviation: 3.9265
As we are using the sample standard deviation to estimate the population standard deviation, we will use the t-statistic.
The degrees of freedom are:
[tex]df=n-1=187-1=186[/tex]
The t-statistic for a 95% interval and 186 degrees of freedom is t=1.9728.
Now, we can calculate the margin of error of the confidence interval:
[tex]E=t \cdot s/\sqrt{n}=1.9728*3.9265/\sqrt{187}=7.7462/13.6748=0.5665[/tex]
The upper and lower bound of the confidence interval are:
[tex]LL=\bar x-t \cdot s/\sqrt{n}=8.636-0.5665=8.0695\\\\UL=\bar x+t \cdot s/\sqrt{n}=8.636+0.5665=9.2025[/tex]
The 95% confidence interval for the true mean is:
[tex]8.0695\leq \mu\leq9.2025[/tex]
b) As the confidence level is equal, the t-value is the same.
But the sample size has to be adjusted to have an margin of error of +/-1.5.
We use the formula for the margin of error:
[tex]E=t \cdot s/\sqrt{n}=1.9728*3.9265/\sqrt{n}=7.7462/\sqrt{n}=1.5\\\\\\n=(\dfrac{7.7462}{1.5})^2=5.1641^2=26.67\approx27[/tex]
This margin of error can be achieved with a sample of size 27 individuals.
