Question

A teacher is interested in her students’ opinion on whether the non-required lecture is effective for them. She flips a coin every time a student walks into lecture and if it lands heads, she asks that student if they find lecture effective and records their answer.

(a) Are there any issues with creating a confidence interval with this sample if our parameter of interest is the proportion of enrolled students who find lecture effective? Explain briefly.

(b) Another lecturer asks a random sample of his students to comment on the efficacy of his lecture and fifty two (52) of the 80 students interviewed reported finding the lecture effective. Create a 98% confidence interval for the proportion of students who find his lecture effective.

Answer 1

a.
yes,
there are any issues with creating a confidence interval with this sample
if our parameter of interest is the proportion of enrolled students who find lecture effective.
A teacher is interested in her students’ opinion on whether the non-required lecture is effective for them.
She flips a coin every time a student walks into lecture and if it lands heads,
she asks that student if they find lecture effective and records their answer.
b.
TRADITIONAL METHOD
given that,
possible chances (x)=52
sample size(n)=80
success rate ( p )= x/n = 0.65
I.
sample proportion = 0.65
standard error = Sqrt ( (0.65*0.35) /80) )
= 0.0533
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.02
from standard normal table, two tailed z α/2 =2.326
margin of error = 2.326 * 0.0533
= 0.124
III.
CI = [ p ± margin of error ]
confidence interval = [0.65 ± 0.124]
= [ 0.526 , 0.774]
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DIRECT METHOD
given that,
possible chances (x)=52
sample size(n)=80
success rate ( p )= x/n = 0.65
CI = confidence interval
confidence interval = [ 0.65 ± 2.326 * Sqrt ( (0.65*0.35) /80) ) ]
= [0.65 - 2.326 * Sqrt ( (0.65*0.35) /80) , 0.65 + 2.326 * Sqrt ( (0.65*0.35) /80) ]
= [0.526 , 0.774]
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interpretations:
1. We are 98% sure that the interval [ 0.526 , 0.774] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 98% of these intervals will contains the true population proportion