2)Let D be the difference between two different measurements of pH. What is the

Question

2)Let D be the difference between two different measurements of pH. What is the standard deviation of D? Remember that D can be both negative and positive!

3)What is the probability that the difference D should be greater than 0.01? That is, what is the probability that two measurements will differ more than 0.01 from each other? Remember that D can be both negative and positive!

4)you have taken four samples from the solution and found these four values: 7.58, 7.49, 7.62 and 7.55.Find a 95% confidence interval

5)How many measurements must we make at least to ensure that the confidence interval does not exceed 0.05?

Explanation / Answer

Solution

Given Xi ~ N(µ, 0.05)

Part (1)

The likelihood that a random observation Xi shall deviate more than 0.01 from the true value

= P(|Xi - µ| > 0.01)

= P(|{(Xi - µ)/0.05}| > (0.01/0.05)

= P(|Z| > 0.2), where Z ~ N(0, 1)

= 0.8414 ANSWER [using Excel Function on N(0, 1)]

Part (2)

Let Xi and Xj be two independent determinations. Then, their difference, D = (Xi – Xj) or (Xj – Xi).

So, variance of D = V(D) = V(Xi) + V(Xj) + V(Xj) + V(Xi) = 4V(X) since Xi and Xj are independent and identically distributed.

So, standard deviation of D = 2SD(Xi) = 2 x 0.05 = 0.1 ANSWER

Part (3)

Since D = (Xi – Xj), Xi ~ N(µ, 0.05) and Xj ~ N(µ, 0.05), D ~ N(0, 0.052)

So, probability that the difference D should be greater than 0.01

= P(|D| > 0.01)

= P(|Z| > (0.01/0.052)

= P[|Z| > {(2)/10}]

= P[|Z| > 0.1414)

= 0.8876 ANSWER

Part (4)

95% confidence interval = {Xbar ± (s/n)(t3, 0.025)} where

Xbar = sample mean, s = sample standard deviation, n = sample size and t3, 0.025 = upper 2.5% point of t-distribution with degrees of freedom = 3

Calculations on the given sample values, 7.58, 7.49, 7.62 and 7.55:

=

0.05

n-1

3

n =

4

sqrtn

2

              

Xbar =

7.56

s =

0.054772256

             tn-1, /2 =

3.182446305

95% CI for

7.56

±

0.087154881

Lower Bound =

7.47284512

Upper Bound =

7.64715488

=

0.05

n-1

3

n =

4

sqrtn

2

              

Xbar =

7.56

s =

0.054772256

             tn-1, /2 =

3.182446305

95% CI for

7.56

±

0.087154881

Lower Bound =

7.47284512

Upper Bound =

7.64715488

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