We Have Numbers Of Free Samples

Table of Content

• Background
• Problem Statement
• Solution
  • 3.1 Normality Test
  • 3.2 State Null and Alternate Hypothesis
  • 3.3 Set the level of significance (α)
  • 3.4 Choose the type of test
  • 3.5 Calculate a test statistic
  • 3.6 Determine P-value, construct Acceptance/Rejection Region.
  • 3.7 Analysis of Boxplot of Method A and B.
  • 3.8 Conclusion
• Referencing

1. Background

The UK’s Committee on Toxicity of Chemicals in Food has launched an investigation about leaching of a chemical called Bisphenol A (BPA), which is a commonly used inert coating material for packaging of food items. It is scientifically proven that, it mimics properties of hormone such as estrogen and is a cause of concern, in order to find out whether there is any leaching of this chemical into the food item packed with the card board coated with this material, an experiment was conducted on two different fruit juice carton packaging methods (Method A and Method B), both using cardboard coated with BPA. 36 cartoons of juice were prepared by method A and 31 cartoons of juice by method B. a total of 67 cartoons of juice was prepared under identical conditions. Further, the fruit juice in these cartoons were analyzed using a technique called High Performance Liquid Chromatography (HPLC) to determine the BPA level in the fruit juice packed using method A and B. The data is presented in table 1.

Table 1: BPA concentrations (µg/L) in fruit juice following packaging by two methods

2. The Problem Statement

Based on the level of BPA found in the fruit juice as presented in table, the Committee wants to know whether there is a significant difference between the two carton packaging processes? If yes, what is the evidence?

Solution:

To find out whether there is any significant difference between the above said methods, we shall perform a Hypothesis testing. At the end of the hypothesis testing we will get statistically significant evidence to say whether there is a significant difference between the two carton packaging processes or not. Following are the steps to be followed:

• Conduct normality test.
• State Null and Alternate Hypothesis.
• Set the Level of Significance.
• Choose the type of test you want to perform as per the sample data.
• Calculate a test statistic.
• Determine P-value, construct Acceptance / Rejection regions.
• Analysis of Box-plot.
• Based on steps 5, 6 and 7, draw a conclusion.

3.1 Normality test

Anderson-Darling normality test is conducted to check the normality of the data, The Figure 1 (a) and Figure 1 (b) shows the results of normality test conducted using Minitab software. From, figure 1(a) and 1(b) the p-values are found to be 0.393 and 0.03 are respectively for method A and B. Since both are higher than 0.05, it can be inferred that, data is normally distributed and hence we can proceed with hypothesis testing.

Figure 1(a): Anderson-Darling Normality test summary report for Method A

Figure 1(b): Anderson-Darling Normality test summary report for Method A

3.2 State Null and Alternate Hypothesis

Our Null Hypothesis, in this problem would be to say, there is no difference in the population mean of BPA level in fruit juice by the methods A and B. Our Alternate Hypothesis would be to say there is a difference in the population mean of BPA level in fruit juice between the method A and B.

Let, µ1 is the population mean of BPA level in the fruit juice packed using Method A, µ2 is the population mean of BPA level in the fruit juice packed using Method B,

H0 is null hypothesis, and H1 is alternate hypothesis

Then mathematically we have,

Null hypothesis             H₀: μ₁ – µ₂ = 0
Alternative hypothesis H₁: μ₁ – µ₂ ≠ 0

3.3 Set the level of significance (α).

It is the measure of how much risk one would prefer to take, if Null hypothesis is rejected even though it is true. We will set level of significance as 5%.

3.4 Choose the type of test you want to perform as per the sample data.

Since, there are two separate samples, we will use 2 sample t-test using Minitab software, it assumed that the population variance of both methods A and B are equal (pooled approach) since it is more superior to unpooled approach [3]. From the mathematical expressions of null and alternate hypothesis it is inferred that, test is a two tailed test.

3.5 Calculate a test statistic.

The results of test statistic computed from Minitab software are presented in figure 2. There are four subsections in the result, the first subsection, presents the symbols used for indicate population mean of

method A and B. The second section presents the descriptive statistics like mean, standard deviation and standard error mean of both methods.

Figure 2: Results of 2 sample t-test computed by Minitab 18 software

The third subsection presents the estimate for the difference between the population mean of BPA level in fruit juice packed by method A and method B. It is important to note that, the difference in population mean is found to be -1.032 and is within the confidence interval (CI) (-1.982 and -0.082) computed at 95% confidence level. Which means that, if the experiments of measuring the BPA level in fruit juice packed by method A & B is conducted 100 times, 95 times the difference in the mean will lie within the CI.

3.6 Determine P-value, construct Acceptance / Rejection regions.

Finally, the fourth subsection in figure 2, presents the results of hypothesis testing, we can observe that degrees of freedom is 64 and is calculated as

= A + B − 2
= 35 + 31 − 2 = 64, Where, A& B are sample size of method A & B.

The t-value computed at 5% is found to be ( ) = −2.17 whereas, at 5% significance level from t-table at 64 DF, we have

Therefore, as indicated in figure 3, t-value lies in the rejection region and hence our Null hypothesis can be rejected.

Figure 3: Region of rejection / acceptance

P-Value: P-value is probability of results being as extreme as actual value, under the assumption that our null hypothesis is correct. In simple words, at 5% significance level, lower is the p-value from 0.05, more surprising is the result and worst is our null hypothesis, therefore in such situation we reject our null hypothesis. P-value is treated as a reliable statistical evidence to reject or accept a null hypothesis.

From figure 2, it can be noted that, the p-value for this example is computed as 0.034, which is less than 0.05. This means we now have enough statistical evidence to reject our null hypothesis and cab be inferred that, there is a significant difference between the two carton packaging processes i.e method A and B.

3.7 Analysis of Boxplot of Method A and B

The figure 4 presents the boxplot for method A and B. It can be observed that, the mean and median value of BPA level in fruit juice packed by method A is lower when compared to that of fruit juice packed by method B. This means, method A of packaging is superior and safer than method B as lower amount of BPA is leaching into fruit juice.

Figure 4: Boxplot of BPA levels in fruit juice packed by Method A and B

3.8 Conclusion

The presented problem was analyzed by 2 sample, 2-tailed t-test, Pooled approach was used while testing, based on the analysis of results presented in steps 5, 6 and 7 following recommendation are made to the committee.

Recommendations:

1. A statistically significant evidence is found during hypothesis testing to state that, there is a significant difference between the population mean of BPA level in fruit juice packed by method A and method B. Hence, there is a significant difference between the two carton packaging processes.
2. The packaging of fruit juice by method A is superior and safer than method B as lower amount of BPA is leaching into fruit juice by method A.