Question

Two methods were used to measure the specific activity (in units of enzyme activity per milligram of protein) of an enzyme. One unit of enzyme activity is the amount that catalyzes the formation of 1 micromole of product per minute under specified conditions. Use an appropriate test or estimation procedure to compare the two methods of measurement. Comment on the validity of any assumptions you need to make. $$ \begin{array}{l|lllll} \text { Method 1 } & 125 & 137 & 130 & 151 & 142 \\ \hline \text { Method 2 } & 137 & 143 & 151 & 156 & 149 \end{array} $$

Short answer

Question: Compare the two methods used to measure the specific activity of an enzyme by using an appropriate test and comment on the validity of the assumptions made. Answer: We used a t-test to compare the means of the two datasets, which resulted in a t-value of -2.98, indicating a significant difference between the two methods. However, the validity of this test is dependent on certain assumptions, such as normal distribution, independent samples, and unequal variances. If these assumptions are incorrect, alternative tests like the Wilcoxon signed-rank test or the Mann-Whitney U test may need to be considered.

Step-by-step solution

Calculate the means of both datasets

First, we need to calculate the means for Method 1 and Method 2. The mean of a dataset is the sum of the values divided by the number of values. Let's denote the mean of Method 1 as \(\overline{x_1}\) and the mean of Method 2 as \(\overline{x_2}\). $$ \overline{x_1} = \frac{125+137+130+151+142}{5} = \frac{685}{5} = 137\\ \overline{x_2} = \frac{137+143+151+156+149}{5} = \frac{736}{5} = 147.2 $$

Calculate the variances and standard deviations of both datasets

Now, we will calculate the variances and standard deviations for both datasets. The variance of a dataset is calculated using the following formula: $$ s^2 = \frac{1}{n-1}\sum_{i=1}^n (x_i - \overline{x})^2 $$ First, let's calculate the variance for Method 1: $$ s_1^2 = \frac{1}{4}\left((125-137)^2+(137-137)^2+(130-137)^2+(151-137)^2+(142-137)^2\right) = 92.5 $$ Now, let's calculate the variance for Method 2: $$ s_2^2 = \frac{1}{4}\left((137-147.2)^2+(143-147.2)^2+(151-147.2)^2+(156-147.2)^2+(149-147.2)^2\right) = 58.2 $$ Standard deviation is simply the square root of the variance: $$ s_1 = \sqrt{92.5} \approx 9.6\\ s_2 = \sqrt{58.2} \approx 7.6 $$

Perform the t-test

We will now use the t-test to compare the means of the two datasets. The t-test formula is: $$ t = \frac{(\overline{x_1} - \overline{x_2}) - 0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} = \frac{(137 - 147.2)}{\sqrt{\frac{92.5}{5} + \frac{58.2}{5}}} \approx -2.98 $$ Here, \(n_1\) and \(n_2\) represent the number of values in Method 1 and Method 2, respectively, and both are equal to 5. Since we have a small sample size (5 in each dataset), we will use a two-tailed t-test, assuming a significance level (alpha) of 5%, which corresponds to a critical value of approximately \(\pm 2.306\). Our t-value of -2.98 is outside of this range, leading to the conclusion that there is a significant difference between the two methods.

Comment on the validity of assumptions

We made a few assumptions while using the t-test: 1. Normal distribution: We assume that both datasets follow a normal distribution, which may not necessarily be the case, especially for small sample sizes. 2. Independent samples: We also assume that the values in Method 1 and Method 2 are independent of each other, meaning that the specific activity of one sample does not influence the specific activity of the other samples. 3. Unequal variances: We have used the t-test assuming unequal variances between the two datasets. If we assume equal variances, we would need to use a different t-test formula (pooled t-test), but this is less flexible and may not be appropriate for this case. With these assumptions, we have concluded that the two methods are significantly different. However, the validity of the test would be compromised if the assumptions were incorrect. For example, if the data were not normally distributed, we may need to consider alternative tests, such as the Wilcoxon signed-rank test or the Mann-Whitney U test.

Key concepts

t-test

The t-test is a statistical method used to determine if there is a significant difference between the means of two groups. It's a valuable tool for hypothesis testing, especially when dealing with small sample sizes. In our exercise, we compared two methods of measuring enzyme activity to see if one method gives significantly different results than the other.

To perform a t-test, you calculate the t-value, which is a ratio of the difference between the group means to the variability of the data. This is done using the formula:

• \[ t = \frac{(\overline{x_1} - \overline{x_2})}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]

Here, \(\overline{x_1}\) and \(\overline{x_2}\) are the means of the two groups, \(s_1^2\) and \(s_2^2\) are their respective variances, and \(n_1\) and \(n_2\) are the sizes of the samples.

By comparing this calculated t-value to a critical value from a t-distribution table, and considering a significance level (like 5%), you can conclude if the difference is significant. In our example, a t-value of -2.98 was obtained, which was outside the range of expected outcomes for equal means, thus indicating a significant difference.

statistical assumptions

When using a t-test, several statistical assumptions must be met to ensure the validity of the results. Understanding these will help you confidently interpret the conclusions.

• **Normal Distribution**: We assume that the data comes from a normally distributed population. For small sample sizes (like 5 in each group), this condition is crucial as departures from normality can lead to incorrect conclusions.


• **Independent Samples**: Each measurement in one group should not influence the measurements in another. This assumption ensures that observations in each group are randomly sampled and independent of each other.


• **Equal or Unequal Variances**: Depending on the type of t-test, assumptions about variance need to be made. An unequal variance t-test (Welch's t-test) does not require the assumption of equal variance, whereas a pooled t-test would.
  
  

Violated assumptions can distort test results. For instance, if data is not normally distributed, alternative tests like the Wilcoxon signed-rank test or the Mann-Whitney U test might be more appropriate.

variance

Variance is a key concept in statistics that describes how much the data points in a set differ from the mean. It's a measure of data spread, indicating how far each number in the set is from the average.

In mathematical terms, variance is calculated by taking the mean of the squared deviations of each point from the mean of the dataset. For a sample, the formula for variance \(s^2\) looks like this:

• \[ s^2 = \frac{1}{n-1}\sum_{i=1}^n (x_i - \overline{x})^2 \]

Where \(n\) is the number of data points, \(x_i\) represents each value, and \(\overline{x}\) is the mean.

In the exercise, we calculated the variance to understand the variability within each method of measuring enzyme activity. It was found that Method 1 had a variance of 92.5, while Method 2 had 58.2, indicating the data from Method 1 was more spread out compared to Method 2.

standard deviation

The standard deviation is the square root of the variance and provides a measure of the spread in the same units as the data. It's a popular statistical tool because it tells us how much data varies from the mean in a straightforward, interpretable way.

If variance tells us about data dispersion, standard deviation puts it into the same units as the data, making it easier to relate the spread directly to the average. Mathematically, if variance is \(s^2\), the standard deviation (\(s\)) is given by:

• \[ s = \sqrt{s^2} \]

In our exercise, the standard deviation helped compare data variability between Method 1 and Method 2 for enzyme activity measurement. With Method 1 having a standard deviation of approximately 9.6, and Method 2 having 7.6, it became clear that readings for Method 1 were more widely spread around the mean compared to Method 2. This insight helps in understanding whether one method consistently produces results close to the mean, improving interpretation of the results.