Question

a simple random sample of 5000 students were selected and asked how likely to for them to maintain a 6 feet distamce in the classrooms if they return to campus. among the sample 5000 , 76% of the students responded that they are somewhat or very likely to keep the physical distance in classrooms
(SHOW All Work) need helpp

1. University is intrested in releasing an interval estimate for the true proportion of students who would be (somewhat or very likely to keep their physical distance). What are the conditions that need to be satisfied in order for us to calculate the confidence interval. Are they satisfied?

2.) Calculate the 90% Confidence interval for the proportion (of students somewhat or very likely to keep their physical distance)?

3.)Interpret the interval calculated in part 2?

4.) Based on the interval you calculated in part 2 we can say more than 70% of penn state students would (somewhat or very likely to keep their physical distance). Explain your anwser?

Answer 1

1.
University is intrested in releasing an interval estimate for the true proportion of students who would be (somewhat or very likely to keep their physical distance).
yes,
the conditions that need to be satisfied in order for us to calculate the confidence interval because standard normal distribution.
2.
TRADITIONAL METHOD
given that,
possible chances (x)=3800
sample size(n)=5000
success rate ( p )= x/n = 0.76
I.
sample proportion = 0.76
standard error = Sqrt ( (0.76*0.24) /5000) )
= 0.006
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
margin of error = 1.645 * 0.006
= 0.01
III.
CI = [ p ± margin of error ]
confidence interval = [0.76 ± 0.01]
= [ 0.75 , 0.77]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possible chances (x)=3800
sample size(n)=5000
success rate ( p )= x/n = 0.76
CI = confidence interval
confidence interval = [ 0.76 ± 1.645 * Sqrt ( (0.76*0.24) /5000) ) ]
= [0.76 - 1.645 * Sqrt ( (0.76*0.24) /5000) , 0.76 + 1.645 * Sqrt ( (0.76*0.24) /5000) ]
= [0.75 , 0.77]
-----------------------------------------------------------------------------------------------
3.
interpretations:
1. We are 90% sure that the interval [ 0.75 , 0.77] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion
4.
yes,
Based on the interval you calculated in part 2 we can say more than 70% of penn state students would (somewhat or very likely to keep their physical distance).
because the confidence interval [ 0.75 , 0.77] is more than 70%.