# Sampling and Data Collection Plan

Topics: Sampling, Sample, Data collection Pages: 6 (595 words) Published: November 9, 2014
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Sampling and Data Collection Plan

Amy Nguyen
QNT/561
November 3, 2014
Dr. Anthony Matias
Sampling and Data Collection Plan
Population and Size
The population would be the customers of Starbucks. The size of the population is 300. Target Population
The target population would be the senior, middle, and young customers of the organization as it would help to find out that how much coffee effect their life. The customers would have more knowledge about the effects of drinking too much coffee daily. Sampling Element

Survey would be used to conduct the research. Following are the questions that would be included: Scalethat would be employed is given below.
SA: Strongly Agree D: Disagree A: Agree SD: Strongly disagree. N: neutral
No.
Questions
SA
A
N
D
SD
1
Coffee is one of the main causes in the amount of coffee assortments you drink daily. 1
2
3
4
5
2
Drinking too much coffee daily can cause health problems
1
2
3
4
5
3
You need to drink at least one coffee per day
1
2
3
4
5
4
You visit Starbucks at least 2 times per week
1
2
3
4
5
5
You think that people should consider avoiding coffee or switching to decaf, especially women who are pregnant, or people who have a hard time controlling their blood pressure or blood sugar. 1
2
3
4
5
6
Heavy daily caffeine use more than 500 to 600 mg a day may cause side effects such as Insomnia Nervousness
Restlessness
Irritability
Stomach upset
Fast heartbeat
Muscle tremors

1
2
3
4
5

Sample size
Forty employees of Starbucks would be used.
Method of sampling
Simple random sampling would be used. The senior, middle, and young customers of the organization would be selected randomly to conduct the survey. Validity and Reliability
Validity and reliability is very much important for a research. If the data would not be reliable or valid then the research would be of no use. To maximize the validity and reliability, a pilot testing would be conducted. Moreover, survey would contain questions that support the research question and help to the expected results (Validity and Reliability, n.d). Protection of human subjects

The respondents would fill the questionnaire anonymously. Their names would not be leaked out. The questionnaire would also contain the statement that “All the responses will be treated confidential and used for the purpose of research only”. No one would get a clue that who participated in the study. Data collection

The data would be collected with the help of the questionnaire. The questionnaire would be send to the respondents through email. Some of the questionnaire would also be given in person by giving to customers at the same time that they want to try something in the store. Physical collection of the data

The questionnaire then would be collected from the respondent after two or three days. Protection and storage of data
Hardcopy of the questionnaire would be stored in one box and it would be labeled with key information so that it could be accessed easily. Moreover, the records would be stored in a locked room so that it does not get in wrong hands. The responses that would be obtained through email would also be protected in an effective manner. The computer would have password and it would not be accessed by anyone. Moreover, there would also be the backup copies of all data to prevent any kind of loss (HEI Records Management, 2007).

References
HEI Records Management. (2007). Retrieved August 10, 2014 fromhttp://tools.jiscinfonet.ac.uk/downloads/bcs-rrs/managing-research-records.pdf Validity and Reliability. (n.d). Retrieved August 10, 2014 fromhttp://www.nationaltechcenter.org/index.php/products/at-research-matters/validity/

Appendix
Calculate the sample size using a 95% confidence level, and a 5% margin of...

Validity and Reliability. (n.d). Retrieved August 10, 2014 fromhttp://www.nationaltechcenter.org/index.php/products/at-research-matters/validity/
Appendix
Calculate the sample size using a 95% confidence level, and a 5% margin of error.
1-α= 1-0.95= 0.05
=α/2
=0.025
Zα/2= Z0.025= 1.96
ME= 5 %= 0.05
n= CI/ ME= 39.2 or 40.