The homogeneity of the chloride level in a water sample from a lake was tested b
ID: 3064598 • Letter: T
Question
The homogeneity of the chloride level in a water sample from a lake was tested by analysing portions drawn from the top and from near the bottom of the lake. The following results in ppm chloride were obtained. 12. Top 26.30 26.43 26.28 26.19 26.49 Bottom 26.22 26.32 26.20 26.11 26.42 (a) (b) (c) Perform-test at the 95% confidence level to determine if the chloride level from the top of the lake is different from that at the bottom Use paired t-test and determine whether there is a significant difference between the top and bottom values at the 95% confidence level Explain why there is a different conclusion drawn from (a) and (b) above.Explanation / Answer
Part a
t critical = 2.3646
t = 1.1078
Since t < t critical, we conclude that there is no significant difference exists at 95% confidence interval.
Detailed results for this test for more explanation are summarized as below: (By using Excel)
Separate-Variances t Test for the Difference Between Two Means
(assumes unequal population variances)
Data
Hypothesized Difference
0
Level of Significance
0.05
Population 1 Sample
Sample Size
5
Sample Mean
26.338
Sample Standard Deviation
0.1207
Population 2 Sample
Sample Size
5
Sample Mean
26.254
Sample Standard Deviation
0.1191
Intermediate Calculations
Numerator of Degrees of Freedom
0.0000
Denominator of Degrees of Freedom
0.0000
Total Degrees of Freedom
7.9985
Degrees of Freedom
7
Standard Error
0.0758
Difference in Sample Means
0.0840
Separate-Variance t Test Statistic
1.1078
Two-Tail Test
Lower Critical Value
-2.3646
Upper Critical Value
2.3646
p-Value
0.3046
Do not reject the null hypothesis
Part b
t critical = 2.7764
t = 12.3851
Since t > t critical, we conclude that there is a significant difference exists at 95% confidence interval.
Detailed results for this test for more explanation are summarized as below: (By using Excel)
Paired t Test
Data
Hypothesized Mean Difference
0
Level of significance
0.05
Intermediate Calculations
Sample Size
5
DBar
0.0840
Degrees of Freedom
4
SD
0.0152
Standard Error
0.0068
t Test Statistic
12.3851
Two-Tail Test
Lower Critical Value
-2.7764
Upper Critical Value
2.7764
p-Value
0.0002
Reject the null hypothesis
Part C
Answer:
Because there are reduced degrees of freedom with the paired t test, the significant difference detected is artificial.
Separate-Variances t Test for the Difference Between Two Means
(assumes unequal population variances)
Data
Hypothesized Difference
0
Level of Significance
0.05
Population 1 Sample
Sample Size
5
Sample Mean
26.338
Sample Standard Deviation
0.1207
Population 2 Sample
Sample Size
5
Sample Mean
26.254
Sample Standard Deviation
0.1191
Intermediate Calculations
Numerator of Degrees of Freedom
0.0000
Denominator of Degrees of Freedom
0.0000
Total Degrees of Freedom
7.9985
Degrees of Freedom
7
Standard Error
0.0758
Difference in Sample Means
0.0840
Separate-Variance t Test Statistic
1.1078
Two-Tail Test
Lower Critical Value
-2.3646
Upper Critical Value
2.3646
p-Value
0.3046
Do not reject the null hypothesis
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