In: Statistics and Probability
You measured dissolved argon concentrations in 20 water streams located at a region. The data is tabulated below. Assume normal distribution.
Observation no |
Concentration mg/L |
Observation no |
Concentration mg/L |
1 |
3.5 |
11 |
3.1 |
2 |
2.7 |
12 |
6.7 |
3 |
5.3 |
13 |
2.3 |
4 |
1.4 |
14 |
4.6 |
5 |
1.7 |
15 |
2.5 |
6 |
0.9 |
16 |
2.3 |
7 |
2.8 |
17 |
4.1 |
8 |
1.6 |
18 |
2.6 |
9 |
8.7 |
19 |
5.3 |
10 |
4.2 |
20 |
8.6 |
(a) Find a 95% confidence interval on the mean dissolved argon concentration.
(b) Find a 95% prediction interval on the dissolved argon concentration for the next stream in the system that will be tested.
(c) Find an interval that will contain 95% of the values of the dissolved argon concenration with 95% confidence.
(d) Explain the difference in three intervals computed in parts (a), (b) and (c).
You are interested in knowing whether these data support a claim that mean dissolved argon concentration is 2 mg/L.
(e) Formulate an appropriate hypothesis testing procedure to investigate this claim.
(f) Test this hypotheses and draw conclusions, using α=0.02. Use P-value for this test.
Answers-
(a) Find a 95% confidence interval on the mean dissolved argon concentration.
Let the 2 limits of the confidence interval be c1 and c2 and the interval [c1,c2] within which the unknown value of the population parameter is expected to lie is called the confidence interval and 95% which is also equal to 1- is also called the confidence coefficient.
Then, z= ( - ) / / Follows N(0,1)
P(-1.96 z 1.96) = 0.95
P(-1.96 ( - ) / / 1.96) =0.95
P( -1.96/ +1.96/) = 0.95
Thus, 1.96/ are the 95% confidence limits for the Mean parameter and the interval ( -1.96/ , +1.96/) is called the 95% confidence interval.
Substituting Values from the attached picture, the 95% confirnce interval is ( 3.75 -1.96 * 2.18/, 3.75 +1.96 * 2.18/) = (2.79, 4.70)
b) Find a 95% prediction interval on the dissolved argon concentration for the next stream in the system that will be tested.
In the same way as above, A prediction interval [l,u] for a future observation X in a normal distribution N(µ,σ2) is written as P[l < X< u]
This can be transformed to Z by subtracting with Mean and dividing with standard deviation.
P [(l - ) / < (x - ) / < (u - ) / ]
Hence,
(l - ) / = -z and (u - ) / =z
l = -z and u = + z
Thus, the prediction interval is written as ( -z , + z)
at 95% confidence, z= 1.96
Substituting the values ( 3.75 - 1.96*2.18 , 3.75 + 1.96*2.18) = (-0.53, 8.02)
Prediction interval will give the interval in which the next observation will fall.
c) Find an interval that will contain 95% of the values of the dissolved argon concenration with 95% confidence.
The interval( -1.96,1.96) is the interval since 95% of the area under the curve falls within this interval.
For a 95% confidence interval covers 95% of the normal curve. The probability of a value falling in the rejection region is less than 0.05. Since it is a symmetric curve, the probability of the value falling on the 2 sides of the rejection region is 0.05/2 = 0.025.
If p = 0.025, the value z* such that P(Z > z*) = 0.025, or P(Z < z*) = 0.975, is equal to 1.96. For a confidence interval with level C = 0.95, the value p is equal to (1-0.95)/2. A 95% confidence interval for the standard normal distribution, then, is the interval (-1.96, 1.96), since 95% of the area under the curve falls within this interval.
d) Explanation already covered in the first 3 parts a, b and c.
e) Testing of Hypothesis
Null Hypothesis H0: The mean dissolved Argan is 2mg/l
Alternative Hypothesis H1: The mean dissolved Argan is not 2mg/L
f) At 0.02 Level of significance.
Z = ( (Mean - ) / )
Cal Z = (3.75 - 2)/0.49
Cal Z = 3.57
Tab Z at 0.02 is 2.326
Thus, Cal Z > Tab Z, this gives us significant evidence to reject the Null Hypothesis. This means that the data does not support the claim that the mean value of the Argan is 2mg/L