Question

In a comparison of two methods for the determination of chromium in rye grass, the following results (mg kg⁻¹ Cr) were obtained:

Method 1: mean = 1.48; standard deviation 0.28.
Method 2: mean = 2.33; standard deviation 0.31.
For each method five determinations were made.
Do these two methods give results having means which differ significantly at 95% confidence?


A. spooled = 0.295
t = 4.56
t > tᵣᵢₜ
With 95% confidence the difference between the two results is significant and the null hypothesis is rejected.
B. spooled = 0.285
t = 4.07
t > tᵣᵢₜ
With 95% confidence the difference between the two results is significant and the null hypothesis is rejected.
C. spooled = 0.295
t = 4.56
t > t
With 95% confidence the difference between the two results is significant and the null hypothesis cannot be rejected.
D. spooled = 0.285
t = 4.07
t > t
With 95% confidence the difference between the two results is significant and the null hypothesis cannot be rejected.

Answer 1

The correct approach involves using a t-test to determine if there is a significant difference between the means of two chromium determination methods. The calculated t-value exceeds the critical value from the t-distribution table at a 95% confidence level, leading to rejection of the null hypothesis and conclusion of significant difference between the two methods.

Step-by-step explanation:

When comparing two methods for the determination of chromium in rye grass, we use a t-test to determine if the means of the two methods are significantly different. With a 95% confidence level, we compare the calculated t-value with the critical t-value (tᵣᵢₜ) from the t-distribution table corresponding to our degrees of freedom (which would be df = n1 + n2 - 2 = 5 + 5 - 2 = 8 for two groups of five measurements each).

In the given scenarios, when we calculate the pooled standard deviation (spooled) and then determine the t-value, we're looking to see if our calculated t exceeds the critical t-value. If the calculated t is higher, we reject the null hypothesis, concluding that there is a significant difference between the two methods. For a two-tailed test at a significance level of 0.05 and 8 degrees of freedom, the critical value from a t-distribution table is approximately 2.306.

Considering the provided choices, only two options correctly state that the null hypothesis should be rejected; however, only one of them gives the correct calculation for spooled and t. Therefore, the correct statement should indicate that we reject the null hypothesis based on the comparison of the calculated t and critical t, affirming the significant difference between the two methods.