Question

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)?

Answer 1

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