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1. The University of Canterbury wants information on whether full-time undergraduate students at Canterbury use certain...

1. The University of Canterbury wants information on whether full-time undergraduate students at Canterbury use certain facilities (e.g. gym, library).

(a) Define the target population in terms of how it differs from the general population.

(b) The proposed method to collect the data is to place students at three places on campus and ask everyone passing by to fill in the questionnaire. It is intended to do this between 12 and 1pm on the Monday to Friday of the last week of semester one (29 May to 2 June 2017). Assume that everyone asked to fill in a questionnaire does so and the results tabulated. Describe how the results from the sample population may not be representative of the target population.

(c) Ultimately the university samples 500 students from the official list of 10,000 full-time undergraduate students at Canterbury. The questionnaire consists of a set of yes / no questions and assume all 500 students responded. If for one yes / no question the percentage of students answering yes was 40%, what is the sample error at 95% confidence level of this sample estimate? Show the formula you have used and your working.

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Answer #1

a) The target population is the full time undergraduate students at Canterbury which is different from the general population which could be the full time undergraduate students in any university. The focus is only on the university students.

b) The proposed method to collect the data between 12 and 1 PM on Monday to Friday of the last week of semester one could potentially leave a section of the students who would not be able to fill in the questionnaire. These students could be those who might have classes at this time OR section of students who don't have any classes at this time & didn't pass through the designated areas. Hence, the sample might not be representative of the target population as not all the students might be able to take the survey.

c) The margin of Error is calculated by the following formula:

z = 1.96 at 95% confidence level, p = 0.40, n = 500

Hence, Sample Error = 0.04294 = 4.29%

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