Answer to Question #170224 in Statistics and Probability for vizky

Question

The telephone company has the business objective of wanting to estimate the proportion of households that would purchase an additional telephone line if it were made available at a substantially reduced installation cost. Data are collected from a random sample of 500 households. The results indicate that 135 of the householdswould purchase the additional telephone line at a reduced installation cost. a. Construct a 99% confidence interval estimate for the population proportion of households that would purchase the additional telephone line. b. How would the manager in charge of promotional programs concerning residential customers use the results in (a)?

Answer 1

(a) The sample proportion is computed as follows, based on the sample size $N=500$ and the number of favorable cases $X=135$

$$\hat{p}=\dfrac{135}{500}=0.27$$

The critical value for $\alpha=0.01$ is $z_c=z_{1-\alpha/2}=2.5758.$  

The corresponding confidence interval is computed as shown below:

$$CI(Proportion)$$

$$=(\hat{p}-z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}, \hat{p}+z_c\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}})$$

$$=(0.27-2.5758\sqrt{\dfrac{0.27(1-0.27)}{500}},$$

$$0.27+2.5758\sqrt{\dfrac{0.27(1-0.27)}{500}})$$

$$=(0.2189, 0.3211)$$

Therefore, based on the data provided, the 99% confidence interval for the population proportion is $0.2189<p<0.3211,$ which indicates that we are 99% confident that the true population proportion $p$  is contained by the interval $(0.2189, 0.3211).$

(b) Multiple the 99% confidence interval estimate computed in (a) by the number of the households in the area polled to obtain a 99% confidence interval of the actual number of households that would purchase an additional telephone line if it were made available at a substantionally reduced installation cost.