In: Statistics and Probability
6. 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 results 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 use the results in (a)?
Solution :
Given that,
n = 500
x = 135
Point estimate = sample proportion =
= x / n = 135/500=0.27
1 -
= 1-0.27 =0.73
At 99% confidence level the z is ,
= 1 - 99% = 1 - 0.99 =
0.01
/ 2 = 0.01 / 2 = 0.005
Z/2
= Z0.005 = 2.576 ( Using z table )
Margin of error = E = Z/2 *
(
(((
* (1 -
)) / n)
= 2.576* (((0.27*0.73)
/ 500)
E = 0.0511
A 99% confidence interval for population proportion p is ,
- E < p <
+ E
0.27-0.0511 < p <0.27+ 0.0511
0.2189< p <0.3211
The 99% confidence interval for the population proportion p is : 0.2189,0.3211