Taste Sensory Discrimination Study:

A Preference for Aquafina over Tap Water

Brenda, you did a very nice job. There were a few issues with your hypotheses and your analysis, but overall the study looks good. Please see my comments throughout your paper.

Grade=91%

Brenda C. Ledford

MGMT 650-9040

Professor Dowell

Ledford 2

November 20, 2011

Purpose of Study

The purpose of this study is to test if a test subject is able to tell the difference between

Aquafina and tap water, and whether or not this ability to discern equals the population

proportion or not. Americans spend almost 4 billion dollars annually on bottled water (Natural

Resources Defense Council, 2011). A four-year study by the Natural Resources Defense Council

(2011) reveals that regulation of the bottled water industry is ambiguous at best and that

government resources confirm that nearly one fourth of bottled water consumed in the

United States (U.S.) is tap water (para. 5). This taste sensory discrimination study offers many

benefits to both individual consumers and businesses that market bottled water. Benefits

include potential economic savings for individuals on the expenditure of bottled water if the

outcome of the study reveals that there is no preference. Businesses on the other hand may

gain valuable knowledge that will assist in both marketing and product improvements.

Although many brands of bottled water exist, this study focuses on Aquafina because

the test subject claims to be able to tell difference between Aquafina and tap water. Aquafina’s

self-proclaimed statement of quality boasts that Aquafina’s purification process meets the U.S.

Food and Drug Administration’s (FDA) standards and that water sources include “rivers, lakes,

streams, ponds, reservoirs, springs, and wells” (Aquafina, 2011, para2). This study utilizes tap

water from the Maquoketa, Iowa municipal water source. The City of Maquoketa recognizes

that the water meets FDA standards for quality (City of Maquoketa, 2011). The City of

Comment [O1]: Good item to test.

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Maquoketa reports that three active public wells serve as the source for its municipal water

supply (City of Maquoketa, 2011).

The null hypothesis for this study states that the test subject cannot tell the difference

between tap water and Aquafina and his skill is random in which he will be able to ‘guess’ right

exactly half of the time based on our trial run study:

H0: p = .50

The alternative hypothesis for this study is that the test subject can tell the difference between

tap water and Aquafina in which case his ability will produce a preference that is either more

than half right or less than:

H1: p ≠ .50

This hypothesis represents the proportion outcomes of the trial study recommended by

Levine, Stephan, Krehbiel, and Berenson (2011) as a guide for population hypothesis testing (p.

298). This hypothesis was developed from the results of the trial study. The trial study yielded

that the trial test subject identified Aquafina 3 out of 6 times. Therefore with respect to the trial

study, n= 6 (total number of events) and x (the number of times the test subject selected

Aquafina) = 3. p = x / n, or p = 3 / 6. p = .50 this is the value represented in H0 (Levine, Stephan,

Krehbiel, & Berenson, 2011, p. 267).

Method

Participants of the Study. The participants of this study must be individuals who drink

water for hydration. The population frame for this study is:

participants who prefer Aquafina

participants who prefer tap water

Comment [O2]: Here, if he is an expert we not only want his guesses to not equal 50%, but we want

him to get MORE than 50% correct (if he gets less than 50% this would not indicate that he can tell the

difference). So, your alternative hypothesis should

be H1>50%. Your null, then should include all other possibilities, so H0<=50%.

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The participants for this study were selected because they were self-proclaimed experts in

being able to identify Aquafina over tap water. The first participant (trial run) was my son , he

was available for the trial run and has expressed a preference for Aquafina over tap water. The

main participant was my husband, who claims to be able to tell the difference between

Aquafina and tap water. This study uses probability sampling because research reveals that

“probability samples should be used whenever possible” (Levine et al., 2011, p.224).

Assignment limitations dictate that the one participant be used for the study, therefore,

random sampling method was selected over systematic, stratified, or cluster sampling methods.

The participants were individuals that the study administrator knows personally and were the

first individuals thought of who met one of the frame criteria and agreed to participate in the

study.

Experiment Design. This study employs a blind sampling design because it was

desirable that the participants not recognize which water they were drinking. The trials were

paired, presenting one of each type of water at the same time and the test subject was told to

identify which one is tap water or which is Aquafina. The cups were marked A or B with the

distribution schedule noted in Table B1 found in Appendix B. The test subject was asked to

taste cup A first then taste B. The Sample Space for this study is:

All Possible Events Probability percentage

Selecting A when Aquafina is A 1 / 4 = .25

Selecting B when Aquafina is B 1/ 4 = .25

Selecting A when tap water is A 1/ 4 = .25

Selecting B when tap water is B 1 /4 = .25

Comment [O3]: This is not probability sampling.

This is basically a judgment sample, which is a non-probability sampling method.

Comment [O4]: Write in past tense since this is a report of a study that has already been completed.

Comment [O5]: Good

Comment [O6]: Why did you choose paired trials over independent trials?

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Total .25 + .25 + .25 + .25 = 1.00

One of each water type will be presented at each event so the probability of selecting

Aquafina during any trial will be 1 out of 2 or 1 / 2 or .50 or 50% probability (based on Levine et

al., 2011, p. 134). Additional factors were taken into consideration to make sure that each trial

remained independent:

To ensure that the random variables in the two water products was as neutral as possible precautions were taken with the tap water. The tap water utilized for this study has a tendency to smell like chlorine when it comes out of the tap, so to avoid this, the tap water was collected days earlier and let it sit out on the counter to let the smell dissipate

To prevent temperature change in the product both products were set out on the same counter for at least a day prior to testing so that they would be the same temperature

To alleviate the tendency in tasting trials of the flavors blurring together research was conducted to evaluate an appropriate palate cleanser. Two options were found to be affordable choices for this study, cracker, or bread .The palate cleanser was chewed between each sample tasting. It was decided to use bread because the crackers available were salted and it was speculated that this would influence the outcome negatively.

Step-by-Step Procedure. The procedure followed for this study was:

1. Gather two identical glasses, and the means to wash the glasses after each use 2. Gather reusable sticky note labels and mark “A” or “B” on two 3. Gather 2 glass liquid measuring cups, such as Pyrex 4. Cut two slices of bread of choice into 1” cubes and place in a sandwich baggie 5. Pour 32oz. of Aquafina into a container and set out to achieve a room

temperature 6. Pour 32oz. of tap water into a container and set out to achieve a room

temperature

Sampling procedure.

1. Pour 1/3 cup of tap water and 1/3 cup of Aquafina into the glasses. Place the labels on the glasses and note in a data table, which is which.

Comment [O7]: Good!

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2. Present the glasses marked “A” and “B” to the test subject and explain the tasting procedure of sipping or drinking all of the first sample of choice and then chewing a cube of bread to cleanse the palate prior to sampling the second sample. The test subject was asked to taste “A” before “B” at each trial.

3. Ask the test subject to indentify “A” or “B” as Aquafina or tap water and record the selection or choice in the data table.

4. Repeat until twelve trials are completed then precede with the Data Analysis steps. For this study it was determined that the 12 trials would span a day to comply with the test subjects time constraints.

The data table for this study may be found within Table B1 in Appendix B of this report. The

table reveals the labeling and product distribution for this study.

Data Analysis

The Null Hypothesis and Alternative Hypothesis both are non-directional statements.

This indicates for a 95% confidence level that the remaining 5% could be split between the

lower and upper tails. Indicating that .025 region of rejection could fall in the lower tail- if the

test subject is not an expert but that the remaining .025 might fall within the upper tail if the

test subject indeed has a skill. Therefore, a two-tailed test is appropriate for this situation.

Although the test subject professes to be an expert in detecting the difference between

Aquafina and tap water, the research indicates that a significant portion of bottled water is tap

water. Therefore, this study administrator felt the outcome of expertise was not certain enough

to be directional. The data analysis implements the Z Test of Hypothesis for the Proportion

because this study is interested in the proportion of preference for tap water among the

population that either is below or above the .50 value stated in H0. The steps in calculating the

data analysis are as follows:

1. Define the hypothesis:

H0: p =.50

Comment [O8]: Actually, your hypotheses should have been directional, indicating that you

should use a one-tailed test and all of the rejection

region is in the upper tail.

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H1: p ≠ .50

2. Determine the sample size, n = 12 . This number was selected because the course text

notes that it is always better to overestimate rather than underestimate the sample

size. Since the text states that a statistic test “will “follo[w] a standardized normal

distribution [if] the number of events of interest and the number of events not of

interest are each at least 5” (Levine et al., 2011, p. 321). That number would have been

10 trials to provide for an even split of 5 for each event of interest. Twelve exceeds this

estimation.

3. Because the Null Hypothesis will be rejected if the population prefers tap water less

than half or more than half of the time a two-tale test is sufficient for determination.

The Z Test of the Hypothesis for the Proportion was selected. The Level of Significance,

α = .05. The Decision Rule for this study is:

Reject the H0 if ZSTAT < - 1.96 or if ZSTAT > + 1.96;

Otherwise, do not reject H0

4. Because the Trial Study indicated that tap water was preferred 3 out of 6 trials, (p) = .50

5. Excel was used to calculate the Sample Proportion by using the respective formula

found within the Excel table in Appendix D, Table D1

6. Then calculate Standard Error, Z Test, Lower and upper Critical Values and the p-Value

following the formulas presented in Appendix D, Table D1

7. The results for the previous values are: Sample Proportion: 1, Standard Error: .1443, Z

Test Statistic: 3.4641, Lower Critical Value: -1.96, Upper Critical Value: 1.96, p-Value:

.000532. These values can be seen in table format within table C1 in Appendix C. Note

Comment [O9]: tail

Comment [O10]: With a one-tailed test, the

critical value is 1.65

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that the Standard Error, Z Test were formatted to 4 decimal places while the Lower and

Upper Critical Values were formatted to two decimal places. The last step was creating

the conditional assessment to either accept or reject the Null Hypothesis as dictated by

the Decision Rule stated previously by implementing the formula within Table D1 in

Appendix D.

Discussion

The calculations of the study data reveal a Z Test Statistic of 3.4641; which is greater

than the upper critical value of + 1.96. This means that the proportion of the study falls within

the upper tail of the distribution that is within a region of rejection. Therefore, the Null

Hypothesis is rejected. Rejection of the Null Hypothesis shows that the test subject exhibits a

high proportion preference for Aquafina and may indeed have expertise to tell the difference

between Aquafina and tap water. This may indicate that the batch of Aquafina utilized was

from a source other than a public well or that the purification processes employed by the

provider masked the taste of well water enough to skew results. The conclusion that can be

drawn from the tests subject’s perfect score is that a sampling error has occurred. Two areas of

sampling errors might be involved: coverage error and/or sampling error. If a coverage error

occurred this would be a reason to examine the frame descriptions and to examine if the test

subject represents the population or if selection bias has occurred (Levine et al., 2011, p. 229).

A sampling error may have occurred because the sampling size was too small (Levine et al.,

2011, p. 230).

Changes that might be considered if the test could be redone include increasing the

number of test subjects to include more random individuals outside the study administrators

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home and increasing the sample size. Increasing the number of test subjects was not detected

at the trial run because the assignment instructions limited the test subjects to one person.

The sample size perhaps was not sufficient because the number of trials used in the trial run

were random. The actual study trials were selected given information in the text. Perhaps if the

trial run had consisted of the ten suggested trials and the actual study had doubled that size the

results would have been different. Additional elements to consider would be to raise the

confidence level to 99%. 95% was selected because the Levine et al. (2011) text indicates this is

the most common confidence level selected. The oversights and uncertainty in this study most

certainly can be attributed to the study administrator’s lack of skill and understanding of

running a study. This was the first such study ever attempted; perhaps future studies would

yield better initial selections and choices.

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References

Aquafina. (2011). Bottled water information statements required under California law. Water

Quality and Information Page. Retrieved from: http://www.aquafina.com/

Levine, D.M., Stephan, D.F., Krehbiel, T.C., & Berenson, M.L. (2011). Statistics for managers

using Microsoft Excel. Boston: Prentice Hall.

Maquoketa, City of. (2011). Water quality report. Chamber of Commerce Home Page. Retrieved

from: http://www.maquoketachamber.com/city/utilityinfo_water.html

Natural Resources Defense Council. (2011). Bottled water pure drink or pure hype? Water Main

Page. Retrieved from: http://www.nrdc.org/water/drinking/bw/exesum.asp

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Appendix A

Table A1

Trial Study Data

Trial # Cup A Cup B Choice

1 Aquafina Tap B

2 Tap Aquafina B

3 Tap Aquafina A

4 Aquafina Tap B

5 Tap Aquafina B

6 Aquafina Tap A

Results:

Aquafina 3/6 = .50 or 50%

Tap Water 3/6= .50 or 50%

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Appendix B

Table B1

Actual Study Data

Trial # Cup A Cup B Choice

1 Aquafina Tap A

2 Tap Aquafina B

3 Aquafina Tap A

4 Aquafina Tap A

5 Tap Aquafina B

6 Tap Aquafina B

7 Aquafina Tap A

8 Tap Aquafina B

9 Aquafina Tap A

10 Aquafina Tap A

11 Tap Aquafina B

12 Tap Aquafina B

Results=

Aquafina 10/10 = 1.0 or 100%

Tap Water 0/10 = 0%

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Appendix C

Table C1

Excel Calculation for the Z Test for the Hypothesis of the Proportion

Z test for the Hypothesis of the Proportion

Data

Null Hypothesis p= 0.5

Level of Significance α= 0.05

Number of items of Interest 12

Sample size 12

Intermediate Calculations

Sample proportion 1

Standard error 0.1443

Z Test Statistic 3.4641

Lower-Tail Test

Lower Critical Value -1.96

Upper Critical Value 1.96

p-Value 0.000532006

reject the null hypothesis

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Appendix D

Table D1

Formula View of the Excel Calculations for the

Z test for the Hypothesis of the Proportion

Z test for the Hypothesis of the Proportion

Data

Null Hypothesis p= 0.5

Level of Significance α= 0.05

Number of items of Interest 12

Sample size 12

Intermediate Calculations

Sample proportion =D6/D7

Standard error =SQRT(D4*(1-D4)/D7)

Z Test Statistic =(D10-D4)/D11

Lower-Tail Test

Lower Critical Value =NORMSINV(D5/2)

Upper Critical Value =NORMSINV(1-D5/2)

p-Value =2*(1-NORMSDIST(ABS(D12)))

=IF(D17<D5,"reject the null hypothesis", "Do not reject the null hypothesis")