Answer:
The 90 % confidence interval for the mean population is (11.176 ; 20.824 )
Rounding to at least two decimal places would give 11.18 , 20.83
Step-by-step explanation:
Mean = x`= 16 miles per hour
standard deviation =s= 4.1 miles per hour
n= 4
[tex]\frac{s}{\sqrt n}[/tex] = 4.1/√4= 4.1/2= 2.05
1-α= 0.9
degrees of freedom =n-1= df= 3
∈ ( estimator t with 90 % and df= 3 from t - table ) 2.353
Using Students' t - test
x`±∈ * [tex]\frac{s}{\sqrt n}[/tex]
Putting values
16 ± 2.353 * 2.05
= 16 + 4.82365
20.824 ; 11.176
The 90 % confidence interval for the mean population is (11.176 ; 20.824 )
Rounding to at least two decimal places would give 11.18 , 20.83
Answer:
[tex]11.18 < \mu <20.82[/tex]
Step-by-step explanation:
From the information given:
A meteorologist who sampled 4 thunderstorms of the sample size n = 16
the average speed at which they traveled across a certain state was 16 miles per hour ; i.e Mean [tex]\bar x[/tex] = 16
The standard deviation [tex]\sigma[/tex] of the sample was 4.1 miles per hour
The objective is to find the 90% confidence interval of the mean.
To start with the degree of freedom df = n - 1
degree of freedom df = 4 - 1
degree of freedom df = 3
At 90 % Confidence interval C.I ; the level of significance will be ∝ = 1 - C.I
∝ = 1 - 0.90
∝ = 0.10
∝/2 = 0.10/2
∝/2 = 0.050
From the tables;
Now the t value when ∝/2 = 0.050 is [tex]t_{\alpha / 2 ,df}[/tex]
[tex]t_{0.050 \ ,\ 3} = 2.353[/tex]
The Margin of Error = [tex]t_{\alpha / 2 ,df} \times \dfrac{s}{\sqrt{n}}[/tex]
The Margin of Error = [tex]2.353 \times \dfrac{4.1}{\sqrt{4}}[/tex]
The Margin of Error = [tex]2.353 \times \dfrac{4.1}{2}[/tex]
The Margin of Error = [tex]2.353 \times 2.05[/tex]
The Margin of Error = 4.82365
The Margin of Error = 4.82
Finally; Assume the variable is normally distributed, the 90% confidence interval of the mean is;
[tex]\overline x - M.O.E < \mu < \overline x + M.O.E[/tex]
[tex]16 -4.82 < \mu < 16 + 4.82[/tex]
[tex]11.18 < \mu <20.82[/tex]
