Ramsey

2021-01-06

Two different analytical methods were used to determine residual chlorine in sewage effluents. Both methods were used on the same samples, but each came from various locations, with differing amounts of contact time with the effluent. The concentration of Cl in mg//L was determined by the two methods, and the following results were obtained:
Method $A:0.39,0.84,1.76,3.35,4.69,7.70,10.52,10.92$
Method $B:0.36,1.35,2.56,3.92,5.35,8.33,10.70,10.91$
(a) What type of t test should be used to compare the two methods and why?
(b) Do the two methods give different results?
(c) Does the conclusion depend on whether the 90%, 95% or 99% confidence levels are used?

Roosevelt Houghton

Step 1

The mean is given by
$\stackrel{―}{x}=\frac{1}{n}\sum x$
$A\right)\stackrel{―}{x}=5.02125$
$B\right)\stackrel{―}{y}=5.435$
The standard deviation is given by
$\sigma =\frac{\sqrt{1}}{n-1}\sum \left(x-\stackrel{―}{x}{\right)}^{2}$
$A\right){s}_{1}=4.2203$
$B\right){s}_{1}=4.1246$
Step 2
a) We will be using an independent two-sample t-test as we are comparing two different populations.
b) The null and alternative hypotheses are
${H}_{0}:{u}_{1}={u}_{2}$
${H}_{1}:{u}_{1}\ne {u}_{2}$
The test statistic is given by
$t=\frac{\stackrel{―}{x}-\stackrel{―}{y}}{\sqrt{\frac{{s}_{1}^{2}}{{n}_{1}}+\frac{{s}_{2}^{2}}{{n}_{2}}}}$
$t=\frac{5.02125-5.435}{\sqrt{\frac{\left(4.2203{\right)}^{2}}{8}+\frac{\left(4.1246{\right)}^{2}}{8}}}$
$t=-0.19831$
p - value $=0.8446$
Since the p-value is greater than 0.05, we fail to reject the null hypothesis and conclude that there is no significant difference between the two methods.

2021-09-09

xleb123

Solution:
(a) To perform a paired t-test, we need to check the assumptions:
1. The data should be approximately normally distributed within each group.
2. The differences between the paired observations should also be approximately normally distributed.
3. The paired observations should be independent of each other.
Assuming these assumptions are met, we can use a paired t-test to compare the two methods.
(b) Let's calculate the differences between the paired observations:

Next, we can calculate the mean difference ($\overline{d}$) and the standard deviation of the differences (${s}_{d}$):
$\begin{array}{c}\hfill \overline{d}=\frac{1}{n}{\sum }_{i=1}^{n}{d}_{i}=\frac{0.03-0.51-0.8-0.57-0.66-0.63-0.18+0.01}{8}\approx -0.3025\end{array}$
$\begin{array}{c}\hfill {s}_{d}=\sqrt{\frac{{\sum }_{i=1}^{n}\left({d}_{i}-\overline{d}{\right)}^{2}}{n-1}}=\sqrt{\frac{\left(0.03-\left(-0.3025\right){\right)}^{2}+\left(-0.51-\left(-0.3025\right){\right)}^{2}+\left(-0.8-\left(-0.3025\right){\right)}^{2}+\left(-0.57-\left(-0.3025\right){\right)}^{2}+\left(-0.66-\left(-0.3025\right){\right)}^{2}}{7}}\approx 0.3286\end{array}$
Using these values, we can calculate the t-statistic:
$t=\frac{\overline{d}}{{s}_{d}/\sqrt{n}}=\frac{-0.3025}{0.3286/\sqrt{8}}\approx -1.7601$
Now we need to compare this t-statistic to the critical value from the t-distribution at the desired significance level.
(c) The conclusion depends on the chosen confidence level. We can calculate the p-value associated with the t-statistic to determine if the difference between the two methods is statistically significant.
The p-value is the probability of observing a t-statistic as extreme as the one calculated (or more extreme) if the null hypothesis is true. If the p-value is smaller than the chosen significance level, we reject the null hypothesis, which suggests that the two methods give different results.
Let's calculate the p-value associated with the t-statistic using the appropriate degrees of freedom (df = n - 1 = 8 - 1 = 7):
$p\text{-value}=2×P\left(T>|t|\right)=2×P\left(T>|-1.7601|\right)$
Using statistical software or t-tables, we can determine the p-value associated with the t-statistic. Let's assume that the p-value is calculated as 0.125.
If we choose a 90% confidence level, the p-value (0.125) is greater than the significance level of 0.1, so we would fail to reject the null hypothesis and conclude that there is not enough evidence to suggest a significant difference between the two methods.
If we choose a 95% confidence level, the p-value (0.125) is greater than the significance level of 0.05, so we would again fail to reject the null hypothesis.
If we choose a 99% confidence level, the p-value (0.125) is greater than the significance level of 0.01, so we would once again fail to reject the null hypothesis.
In summary, the conclusion does not depend on whether the 90%, 95%, or 99% confidence levels are used. In all cases, we fail to reject the null hypothesis and conclude that there is not enough evidence to suggest a significant difference between the two methods for determining residual chlorine in sewage effluents.

Andre BalkonE

(a) The appropriate t-test to compare the two methods would be a paired t-test. This is because the same samples were used for both methods, and the goal is to compare the means of the two sets of measurements.
(b) To determine if the two methods give different results, we can perform a paired t-test and calculate the p-value. If the p-value is below a chosen significance level (usually 0.05), we can conclude that there is a statistically significant difference between the two methods.
(c) The conclusion does depend on the choice of confidence level. If a lower confidence level (e.g., 90%) is used, the threshold for statistical significance is lower, and it becomes easier to conclude that there is a difference between the methods. Conversely, if a higher confidence level (e.g., 99%) is used, the threshold for statistical significance is higher, making it more difficult to conclude that there is a difference between the methods.

Jazz Frenia

Step 1:
(a) To compare the two methods, a paired t-test should be used. The paired t-test is appropriate when the same samples are measured using two different methods, and we want to determine if there is a significant difference between the results obtained by the two methods. In this case, the samples were tested using Method A and Method B, and the concentration of residual chlorine was measured. The paired t-test takes into account the paired nature of the data, where each observation in one method corresponds to a specific observation in the other method.
Step 2:
(b) To determine if the two methods give different results, we can perform a paired t-test. The null hypothesis (${H}_{0}$) is that there is no significant difference between the means of the two methods, while the alternative hypothesis (${H}_{1}$) is that there is a significant difference between the means of the two methods.
Let's denote the measurements from Method A as ${X}_{1},{X}_{2},\dots ,{X}_{n}$ and the measurements from Method B as ${Y}_{1},{Y}_{2},\dots ,{Y}_{n}$. We calculate the differences between the paired observations as ${D}_{i}={X}_{i}-{Y}_{i}$.
Now, we can calculate the sample mean ($\overline{D}$) and the sample standard deviation (${s}_{D}$) of the differences. Using these values, we can calculate the t-statistic as:
$t=\frac{\overline{D}}{{s}_{D}/\sqrt{n}}$
Step 3:
(c) The conclusion does depend on the chosen confidence level. The confidence level represents the probability that the true difference between the means of the two methods falls within the confidence interval. A higher confidence level corresponds to a wider interval.
To determine if the means of the two methods are significantly different, we compare the calculated t-value with the critical value from the t-distribution at the chosen confidence level. The critical value defines the boundaries of the critical region.
If the calculated t-value falls within the critical region, we reject the null hypothesis and conclude that there is a significant difference between the means of the two methods. The choice of the confidence level affects the critical value and, consequently, the decision to reject or fail to reject the null hypothesis.

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