(Hypothetical) You are conducting research on the prevalence of severe binge dri
ID: 3071761 • Letter: #
Question
(Hypothetical) You are conducting research on the prevalence of severe binge drinking among college students. You define severe binge drinking as consuming 5 or more alcoholic beverages at a single sitting. You ask the question: “On how many days in the past two weeks have you consumed 5 or more drinks at one time? The answers can range from 0 to 14. You collect data on a random sample of 900 students.
The average for this sample in 1.37 with a standard deviation of .80. Construct a 95% confidence interval for the average binge drinking score in the population.
Another student who reviewed your figures, says your math is correct, but using the normal distribution isn’t correct because the binge drinking scores are highly skewed, with some students binging very frequently, and the majority of students never binge drinking at all. Is she correct that it is inappropriate to calculate a confidence interval for this research question? (30%)
Explanation / Answer
The formula for estimation is:
= M ± t(sM)
where:
M = sample mean
t = t statistic determined by confidence level
sM = standard error = (s2/n)
Sample Mean (M): 1.37
Sample Size (n): 900
Standard Deviation (s): 0.80
Confidence Level: 95%
Calculation
M = 1.37
t = 1.96
sM = (0.82/900) = 0.03
= M ± t(sM)
= 1.37 ± 1.96*0.03
= 1.37 ± 0.0523
CI [1.3177, 1.4223].
Yes she is correct tht f the distribution is nor normal hence we cannot use confidence interval for this research question, because it is only valid if there is normal distribution.
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