Question

In: Statistics and Probability

A Cellphone provider has the business objective of wanting to estimate the proportion of subscribers who...

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)?

Solutions

Expert Solution

ANSWER:

  • Consider random variable x as the number of subscribers who would upgrade to a new cell phone with improved features.
  • The 99%Confidence interval estimate of Population Proportion (π) of subscribers that would upgrade to new cell phone with reduced cost is computed with the formula.

P ± Zα/2 √P (1-p)/n

OR,

(P- Zα/2   √ P (1-P)/n,   P+ Zα/2 √P (1-P)/n

  • Here
  • X =Number of item in the sample having a certain character
  • n =Sample size
  • π = Proportion of item in the population having a certain character
  •   Zα/2 = Proportion of items in the upper tail probability of standardized normal distribution.
  • p = Proportion of items in the sample having a certain character

= 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.

               


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