In: Statistics and Probability
A Cellphone provider has the business objective of wanting to estimate the proportion of subscribers who would upgrade to a new cellphone with improved features if it were made available at a substantially reduced cost. Data are collected from a random sample of 500 subscribers. The result indicate that 135 of the subscribers would upgrade to a new cellphone at a reduced cost.
a. Construct a 99% confidence interval estimate for the population proportion of subscribers that would upgrade to a new cellphone at a reduced cost.
b. How would the manager in charge of promotional programs concerning residential customers use the results in (a)?
ANSWER:
P ± Zα/2 √P (1-p)/n
OR,
(P- Zα/2 √ P (1-P)/n, P+ Zα/2 √P (1-P)/n
= X/n
=135/500
=.27
A):
From the data values are given are as follows,
X=135
N=500
P=.27
Also Zα/2 = Critical value of the upper tail probability standardized normal distribution
For 99% level of confidence, Zα/2=2.58
So,
Zα/2= Z 0.001/2
=Z 0.005
=2.58
The confidence interval of population proportion (π) is determined by substituting values as follows.
The lower limit is
P- Zα/2 √ P(1-P)/n=.27-(2.58) √. 27(1-2.7)/500
=0219
The upper limit is
P+ Zα/2 √ P (1-P)/n
=.27+ (2.58) √. 27(1-.27)/500
=0.322
Thus the 99% confidence interval estimate of population proportion (π) of subscribers that would up grate to new cell phone with reduced cost is (0.219,0.322)
B) :
The manager in charge of promotional programs would conclude that proportion of subscribers who would upgrade to a new cell phone with improved feature if it were available at substantially reduced cost would lie within 0.22 and 0.32 with 99% confidence.