Respuesta :
Answer: -8.16 to 15.84
Step-by-step explanation: Confidence Interval is an interval in which we are a percentage sure the true mean is in the interval.
A confidence interval for a difference between two means and since sample 1 and sample 2 are under 30, will be
[tex]x_{1}-x_{2}[/tex] ± [tex]t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }[/tex]
where
x₁ and x₂ are sample means
t is t-score
[tex]S_{p}[/tex] is estimate of standard deviation
n₁ and n₂ are the sample numbers
The estimate of standard deviation is calculated as
[tex]S_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2} }[/tex]
where
s₁ and s₂ are sample standard deviation of each sample
Degrees of freedom is:
[tex]df=n_{1}+n_{2}-2[/tex]
df = 12 + 9 - 2
df = 19
Checking t-table, with 90% Confidence Interval and df = 19, t = 1.729.
The mean and standard deviation for 12 unlogged forest plots are 17.5 and 3.53, respectively.
The mean and standard deviation for 9 logged plots are 13.66 and 4.5, respectively.
Calculating estimate of standard deviaton:
[tex]S_{p}=\sqrt{\frac{(12-1)(3.53)^{2}+(9-1)(4.5)^{2}}{12+9-2} }[/tex]
[tex]S_{p}=\sqrt{\frac{299.07}{19} }[/tex]
[tex]S_{p}=[/tex] 15.74
The difference between means is
[tex]x_{1}-x_{2}[/tex] = 17.5 - 13.66 = 3.84
Calculating the interval:
[tex]t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }[/tex] = [tex]1.729.15.74.\sqrt{\frac{1}{12} +\frac{1}{9} }[/tex]
[tex]t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }[/tex] = [tex]27.21\sqrt{\frac{21}{108} }[/tex]
[tex]t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }[/tex] = [tex]27.21\sqrt{0.194}[/tex]
[tex]t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }[/tex] = 12
Then, interval for the difference in mean is 3.84 ± 12, which means the interval is between:
lower limit: 3.84 - 12 = -8.16
upper limit: 3.84 + 12 = 15.84
The interval is from -8.16 to 15.84.
