• A Crot solution of 0.175 mg/L was prepared and analyzed eight times
over the course of several days. The results were 0.195, 0.167, 0.178,
0.151, 0.176, 0.155, 0.154, 0.164 mg/L.
i. Determine the mean, standard deviation, confidence interval at 95%
two-tail.
ii. Calculate the method detection limit (MDL)
iii. What is the estimated practical quantitation limit (PQL)?
iv. Is it accurate to measure samples containing Cr6+ around 0.1 mg/L?

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  1. The mean for these statistic is equal to 0.1675.
  2. The standard deviation for these statistic is equal to 0.0149.
  3. The confidence interval at 95% is 0.1675 ± 0.0104.
  4. The method detection limit (MDL) is equal to 0.035 mg/L.
  5. The estimated practical quantitation limit (PQL) is equal to 29.8 mg/L.
  6. Yes, it is accurate to measure samples containing Cr⁶⁺ around 0.1 mg/L because it's greater than 0.0875 mg/L.

How to calculate the mean of a statistic?

Mathematically, the mean for these statistic can be calculated by using the following formula:

Mean = [F(x)]/n

For the total number of data, we have:

F(x) = 0.195 + 0.167 + 0.178 + 0.151 + 0.176 + 0.155 + 0.154 + 0.164

F(x) = 1.34.

Substituting the parameters into the formula, we have:

Mean = [F(x)]/n

Mean = [1.34]/8

Mean = 0.1675.

How to calculate the standard deviation?

Mathematically, the standard deviation for these statistic can be calculated by using the following formula:

σ = √(x - μ)²/(n - 1)

(x - μ)² = [(0.195 - 0.1675)² + (0.167 - 0.1675)² + (0.178 - 0.1675)² + (0.151 - 0.1675)² + (0.176 - 0.1675)² + (0.155 - 0.1675)² + (0.154 - 0.1675)² + (0.164 - 0.1675)²]

(x - μ)² = [0.0275² + 0.0005² + 0.0105² + 0.00165² + 0.0085² + 0.0125² + 0.0135² + 0.0035²].

(x - μ)² = [0.0007563 + 0.00000025 + 0.0001103 + 0.0002723 + 0.00007225 + 0.0001563 + 0.0001823 + 0.0001225].

(x - μ)² = 0.001562.

Substituting the parameters into the formula, we have:

σ = √0.001562/(8 - 1)

σ = √0.001562/7

σ = √0.00022314285714286

Standard deviation, σ = 0.0149.

How to calculate the confidence interval?

First of all, we would calculate the standard error of the mean (SEM) as follows:

SEM = σ/√n

SEM = 0.0149/√8

SEM = 0.0053.

Alpha, α = 1 - 95/100

Alpha, α = 1 - 0.95

Alpha, α = 0.05

Critical probability (p*) = 1 - α/2

Critical probability (p*) = 1 - 0.05/2

Critical probability (p*) = 0.975.

From the z-table, the z-score is given by:

Zα/2 = 1.960.

Margin of error is given by:

Margin of error, E = 1.960 × 0.0053

Margin of error, E = 0.0104.

Therefore, the confidence interval at 95% is given by:

p - E < p < p + E

0.1675 - 0.0104 < p < 0.1675 + 0.0104

Confidence interval = 0.1675 ± 0.0104.

How to calculate the method detection limit (MDL)?

MDL = Student’s t value × Standard deviation.

MDL = 2.365 × 0.0149

MDL = 0.035 mg/L.

How to calculate the estimated practical quantitation limit (PQL)?

PQL = 100/α × Standard deviation.

PQL = 100/0.05 × 0.0149

PQL = 29.8 mg/L.

How to determine the accuracy?

In order to determine the accuracy, we would calculate the lower limit from the method detection limit (MDL) as follows:

Lower limit = 2.5 × MDL

Lower limit = 2.5 × 0.035

Lower limit = 0.0875 mg/L.

In conclusion, we can infer and logically deduce that it is accurate to measure samples containing Cr⁶⁺ around 0.1 mg/L because it's greater than the "lower limit" of 0.0875 mg/L.

Read more on confidence interval here: brainly.com/question/24156808

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