Question

TV advertising agencies face growing challenges in reaching audience members because viewing TV programsvia digital streaming is increasingly popular. The Harris poll reported on November 13, 2012 that 53% of2343 American adults surveyed said they have watched digitally streamed TV programming on some typeof device.

a) What is the distribution of the sample proportion for this scenario? Justify your answer.

b) Calculate and interpret a confidence interval at the 90% confidence level for the proportion of all adultAmericans who have watched streamed programming.

c) Someone claimed that fewer than 50% of American adults watch digitally streamed TV programming. Does the CI support this claim or contradict it? Justify your answer.

d) What sample size would be required for the width of a 90% confidence interval for π(the population proportion), to be at most 0.05,irrespective of the value of p(remember to use π=0.5 to calculate the standard error)?e) To what population can these results be extrapolated?

Answer 1

a.
the distribution of the sample proportion for this scenario is standard normal distribution
b.
interpret a confidence interval at the 90% confidence level for the proportion of all adultAmericans who have watched streamed programming
TRADITIONAL METHOD
given that,
possible chances (x)=1241.79
sample size(n)=2343
success rate ( p )= x/n = 0.53
I.
sample proportion = 0.53
standard error = Sqrt ( (0.53*0.47) /2343) )
= 0.01
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 * 0.01
= 0.017
III.
CI = [ p ± margin of error ]
confidence interval = [0.53 ± 0.017]
= [ 0.513 , 0.547]
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DIRECT METHOD
given that,
possible chances (x)=1241.79
sample size(n)=2343
success rate ( p )= x/n = 0.53
CI = confidence interval
confidence interval = [ 0.53 ± 1.645 * Sqrt ( (0.53*0.47) /2343) ) ]
= [0.53 - 1.645 * Sqrt ( (0.53*0.47) /2343) , 0.53 + 1.645 * Sqrt ( (0.53*0.47) /2343) ]
= [0.513 , 0.547]
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interpretations:
1. We are 90% sure that the interval [ 0.513 , 0.547] 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
c.
Someone claimed that fewer than 50% of American adults watch digitally streamed TV programming.
no,
the CI support this claim or contradict it.
because the fewer than 50% of American adults watch digitally streamed TV programming are not in the range of confidence interval.
d.
given data,
sample proportion =0.5
margin of error =0.05
confidence level=90%
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.1 is = 1.645
Sample Proportion = 0.5
ME = 0.05
n = ( 1.645 / 0.05 )^2 * 0.5*0.5
= 270.603 ~ 271          
e.
No,
population can these results be extrapolated
The simplest method of extrapolation is to compute the average annual number by which the population has increased from one census to the next,
and to add an equal number for every year which has elapsed since the last census.