A survey of 200 students is selected randomly on a large university campus. They
ID: 3135053 • Letter: A
Question
A survey of 200 students is selected randomly on a large university campus. They are asked if they use a laptop in class to take notes. Suppose that based on the survey, 70 of the 200 students responded "yes".
8a) What is the value of the sample proportion, p ?
8b) What is the standard error of the sample proportion?
8c) Construct an approximate 95% confidence interval for the true proportion (pi) by taking +/- 2 SEs (standard errors) from the sample proportion.
8d) What would be the minimum sample size for this data with a desire sampling error of 2% (that is, e = +/- 0.02) at a 95% confidence?
Explanation / Answer
a)
Here,
p^ = x/n = 70/200 = 0.35 [ANSWER]
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b)
Here,
n = 200
p = 0.35
Hence,
s = standard deviation = sqrt(p(1-p)/n) = 0.033726844 [ANSWER]
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c)
Note that
p^ = point estimate of the population proportion = x / n = 0.35
Also, we get the standard error of p, sp:
sp = sqrt[p^ (1 - p^) / n] = 0.033726844
Now, for the critical z,
alpha/2 = 0.025
Thus, z(alpha/2) = 2
Thus,
Margin of error = z(alpha/2)*sp = 0.067453688
lower bound = p^ - z(alpha/2) * sp = 0.282546312
upper bound = p^ + z(alpha/2) * sp = 0.417453688
Thus, the confidence interval is
( 0.282546312 , 0.417453688 ) [ANSWER]
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d)
Note that
n = z(alpha/2)^2 p (1 - p) / E^2
where
alpha/2 = 0.025
Using a table/technology,
z(alpha/2) = 1.959963985
Also,
E = 0.02
p = 0.35
Thus,
n = 2184.829704
Rounding up,
n = 2185 [ANSWER]
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