Statistics which t-test to use etc..

Question

Q1 A liquid phase reaction: A → 2R was performed in two different reactors and the following
results were obtained in repeated experiments:
Reactor 1: Conversion of A (in %): 57.2, 53.1, 59.8, 58.3, 62.4, 53.2
Reactor 2: Conversion of A (in %): 52.2, 53.4, 55.8, 57.0, 51.0
(a) Is there enough evidence to conclude with 95% confidence that the two reactors give different
conversion of A?
(b) Should the value of 62.4 be rejected as an outlier? Use two different tests to clarify this issue
and compare your findings.
(c) If the value of 62.4 should be retained, estimate with 95% confidence how large this value
would have to be before it could be rejected as an outlier. Use both tests and express this
threshold value using 3 significant figures.

Answer 1

(a) To determine whether the two reactors result in different conversions of A, we need to perform t-test comparing the means of two independent samples. Since the sample sizes are small (n₁ = 6 and n₂ = 5) and the variances may be unequal, we use Welch’s t-test.

Null hypothesis (H₀) : the mean conversions are equal

Alternative hypothesis (H₁): the mean conversions differ

The sample mean and standard deviation for Reactor 1 are 57.33% and 3.38, respectively, while for Reactor 2 they are 53.88% and 2.33. The t-statistic is calculated as:

t = (x̄₁ - x̄₂) / sqrt(s₁² / n₁ + s₂² / n₂)

= (57.33 - 53.88) / sqrt(3.38² / 6 + 2.33² / 5)

≈ 1.996

With approximate degrees of freedom of 9, the critical t-value at a 95% confidence level is 2.262. Since 1.996 < 2.262, we fail to reject the null hypothesis. Therefore, there is not enough evidence at the 95% confidence level to conclude that the two reactors yield significantly different conversions of A.

(b) To assess whether the value 62.4% in Reactor 1 is an outlier, we apply two common statistical tests: Grubbs’ test and Dixon’s Q test. For Grubbs’ test, we calculate the G statistic:

G = |x - x̄| / s = |62.4 - 57.33| / 3.38 ≈ 1.50

The critical value for G at α = 0.05 with n = 6 is 1.887. Since 1.50 < 1.887, the value is not an outlier. For Dixon’s Q test, we use the sorted data and compute:

Q = (x_n - x_{n-1}) / (x_n - x_1) = (62.4 - 59.8) / (62.4 - 53.1) ≈ 0.279

The critical Q value for n = 6 at α = 0.05 is 0.625. Again, since 0.279 < 0.625, we do not reject the value as an outlier. Both tests indicate that the value of 62.4 should be retained.

(c) If the 62.4% value is to be retained, we can estimate how large a value would need to be before it could be considered an outlier using both Grubbs’ and Dixon’s tests. For Grubbs’ test, solving for the threshold:

|x - x̄| = G_crit * s = 1.887 * 3.38 ≈ 6.38

x_threshold = x̄ + 6.38 = 57.33 + 6.38 = 63.7

Thus, a value would need to exceed 63.7% to be rejected under Grubbs’ test. For Dixon’s test, solving for x:

Q = (x - 59.8) / (x - 53.1) = 0.625

→ x ≈ 70.9

So, under Dixon’s test, a value must exceed approximately 70.9% to be considered an outlier. Therefore, the thresholds for rejection using Grubbs’ and Dixon’s tests are 63.7% and 70.9%, respectively (both rounded to three significant figures).