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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

Expert verified
Yes, there is a significant difference between the two methods of measurement. The absolute value of the calculated t-value (3.305) is greater than the critical t-value (2.306), so we reject the null hypothesis and conclude that there is a significant difference between the two methods.

Step by step solution

01

Calculate Mean and Standard Deviation of Each Method

First, let's calculate descriptive statistics for each method in the data. We'll find the mean (µ) and standard deviation (σ) of both datasets. Here are the numbers for each method: $$ \text { Method } 1: 125, 137, 130, 151, 142 $$ $$ \text { Method } 2: 137, 143, 151, 156, 149 $$ Using the formulas for mean and standard deviation: $$ \mu = \frac{\sum x}{n} $$ $$ \sigma = \sqrt{\frac{\sum(x - \mu)^2}{n}} $$ We find: Mean for Method 1: µ1 = 137 Standard Deviation for Method 1: σ1 = 9.23 Mean for Method 2: µ2 = 147.2 Standard Deviation for Method 2: σ2 = 6.76
02

Perform a t-test

Now we can perform a t-test to compare the means of these two samples. The t-test formula is: $$ t = \frac{(\mu_1 - \mu_2)}{(\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2})^{0.5}} $$ We have: Mean for Method 1: µ1 = 137 Standard Deviation for Method 1: σ1 = 9.23 Mean for Method 2: µ2 = 147.2 Standard Deviation for Method 2: σ2 = 6.76 Sample sizes (both methods): n1 = n2 = 5 Plugging these values into the t-test formula, we find: t = -3.305
03

Find the Critical t-value

Next, let's find the critical t-value for our test. For a two-tailed test with a significance level α = 0.05 and degrees of freedom (df) = n1 + n2 - 2 = 8, we find the critical t-value from a t-distribution table: t-critical ≈ 2.306
04

Compare t-value to Critical t-value and Draw Conclusions

Since the absolute value of our calculated t-value (|-3.305| = 3.305) is greater than the critical t-value (2.306), we reject the null hypothesis and conclude that there is a significant difference between the two methods of measurement. As for the assumptions we made for this analysis, the most critical one is that the data are normally distributed and have equal variances. These assumptions are necessary for applying a t-test. While we don't have much information on the actual data distribution, the limited number of data points (5 per method) may not give us enough information to ensure that the assumptions of normality and equal variances are reasonable. Therefore, a non-parametric test could also be considered in cases where the assumptions of a t-test are not met. However, under these assumptions, our t-test shows a significant difference between the two methods.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Descriptive Statistics
Descriptive statistics involves summarizing and organizing the data so it can be understood clearly. In scientific research like the measurement of enzyme activity, this typically includes calculating the mean and standard deviation of each set of measurements.

The mean represents the average value, showing the central tendency of the data, while the standard deviation indicates the spread of data or variability around the mean. High standard deviation means data is spread widely from the mean and vice versa. For instance, in the given problem, the mean (µ) calculation is based on the formula \( \mu = \frac{\sum x}{n} \) where 'x' represents each value in the data set and 'n' is the number of values.

Standard deviation (σ) is calculated using \( \sigma = \sqrt{\frac{\sum(x - \mu)^2}{n}} \) which considers how each data point deviates from the mean. The descriptive statistics provide a foundation for further analysis and understanding the behavior of the data, especially when verifying assumptions of statistical tests like the t-test.
Normal Distribution Assumption
Many statistical tests, including the t-test, are based on the assumption that the data follows a normal distribution, also referred to as the bell curve. This distribution assumes that most of the data points are centered around the mean, with symmetrical tails on either side.

In practice, the normal distribution assumption allows us to use probability theory to make inferences about the population from which our sample is drawn. For small sample sizes, it might not be easy to prove normality, and the t-test could be less reliable. It is essential to interpret t-test results with caution, especially if the sample size is small, like in the exercise given, where we have 5 measurements for each method. If the assumption is violated, the test's conclusion might not be valid, suggesting that non-parametric alternatives should be considered.
Hypothesis Testing
Hypothesis testing is a method used to determine if there is a significant effect based on sample data. In the context of comparing enzymatic activity using different methods, we begin by stating a null hypothesis (often that there is no difference between methods) and an alternative hypothesis (typically that there is a difference).

Using a t-test in hypothesis testing aims to compare means from two different groups to see if they are statistically different from each other. The calculation for this t-test involves the means, standard deviations, and sample sizes of both groups and is expressed as \( t = \frac{(\mu_1 - \mu_2)}{(\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2})^{0.5}} \). The result (t-value) is then compared against a critical value based on degrees of freedom and a significance level (usually 0.05), which allows us to either reject or fail to reject the null hypothesis.
Enzyme Activity Measurement
Measuring enzyme activity is crucial in biochemical research and diagnostics. It involves calculating the rate at which a substrate is converted into a product by an enzyme under specific conditions. One unit of enzyme activity is typically defined as the amount that catalyzes the formation of 1 micromole of product per minute.

In the exercise, two methods are used to measure enzyme activity, and the measurements are compared statistically to determine if there is a significant difference in performance between the two methods. With the use of descriptive statistics and a t-test, analysis of the measured enzyme activity provides insights not only into the enzyme's catalytic capabilities but also into the methods' reliability and accuracy.

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Most popular questions from this chapter

Car manufacturers try to design the bumpers of their automobiles to prevent costly damage in parking-lot type accidents. To compare repair costs of front versus back bumpers for several brands of cars, the cars were subject to a front and rear impacts at \(5 \mathrm{mph},\) and the repair costs recorded. \({ }^{15}\) \begin{tabular}{lrr} Vehicle & Front & Rear \\ \hline WW Jetta & \(\$ 396\) & \(\$ 602\) \\ Daewoo Nubira & 451 & 404 \\ Acura 3.4 RL & 1123 & 968 \\\ Dodge Neon & 687 & 748 \\ Nissan Sentra & 583 & 571 \end{tabular} Do the data provide sufficient evidence to indicate that there is a significant difference in average repair costs for front versus rear bumper repairs costs? Test using \(\alpha=.05 .\)

A paired-difference experiment was conducted to compare the means of two populations: Pairs Population 2 3 4 5 \(\begin{array}{llllll}1 & 1.3 & 1.6 & 1.1 & 1.4 & 1.7 \\\ 2 & 1.2 & 1.5 & 1.1 & 1.2 & 1.8\end{array}\) a. Do the data provide sufficient evidence to indicate that \(\mu_{1}\) differs from \(\mu_{2}\) ? Test using \(\alpha=.05\). b. Find the approximate \(p\) -value for the test and interpret its value. c. Find a \(95 \%\) confidence interval for \(\left(\mu_{1}-\mu_{2}\right) .\) Compare your interpretation of the confidence interval with your test results in part a. d. What assumptions must you make for your inferences to be valid?

Eight obese persons were placed on a diet for 1 month, and their weights, at the beginning and at the end of the month, were recorded: Weights \begin{tabular}{ll} Subjects & Initial \\ \hline \end{tabular} Final \(\begin{array}{lll}1 & 310 & 263 \\ 2 & 295 & 251 \\ 3 & 287 & 249 \\ 4 & 305 & 259 \\ 5 & 270 & 233 \\ 6 & 323 & 267 \\ 7 & 277 & 242 \\ 8 & 299 & 265\end{array}\) Estimate the mean weight loss for obese persons when placed on the diet for a 1 -month period. Use a \(95 \%\) confidence interval and interpret your results. What assumptions must you make so that your inference is valid?

Under what assumptions can the \(F\) distribution be used in making inferences about the ratio of population variances?

Refer to Exercise \(10.7,\) in which we measured the dissolved oxygen content in river water to determine whether a stream had sufficient oxygen to support aquatic life. A pollution control inspector suspected that a river community was releasing amounts of semi treated sewage into a river. To check his theory, he drew five randomly selected specimens of river water at a location above the town, and another five below. The dissolved oxygen readings (in parts per million) are as follows: \begin{tabular}{l|lllll} Above Town & 4.8 & 5.2 & 5.0 & 4.9 & 5.1 \\ \hline Below Town & 5.0 & 4.7 & 4.9 & 4.8 & 4.9 \end{tabular} a. Do the data provide sufficient evidence to indicate that the mean oxygen content below the town is less than the mean oxygen content above? Test using \(\alpha=05 .\) b. Suppose you prefer estimation as a method of inference. Estimate the difference in the mean dissolved oxygen contents for locations above and below the town. Use a \(95 \%\) confidence interval.

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