Question

In a survey of 300 random households in a particular village, mother's have been asked if they used cellphone to communicate there are 76 who said YES, Use 90%, 95% and 99% confidence to estimate the proportion of all mothers who use the cellphone to communicate

Answer 1

The sample proportion is computed as follows, based on the sample size $N = 300$ and the number of favorable cases $X = 76:$

$$\hat{p}=\dfrac{X}{N}=\dfrac{76}{300}\approx0.253333$$

1. The critical value for $\alpha = 0.10$ is $z_c = z_{1-\alpha/2} = 1.6449.$

The corresponding confidence interval is computed as shown below:

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

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

$$=(\dfrac{76}{300}-1.6449\sqrt{\dfrac{\dfrac{76}{300}(1-\dfrac{76}{300})}{300}},$$

$$\dfrac{76}{300}+1.6449\sqrt{\dfrac{\dfrac{76}{300}(1-\dfrac{76}{300})}{300}})$$

$$=(0.212,0.295)$$

Therefore, based on the data provided, the 90% confidence interval for the population proportion is $0.212 < p < 0.295,$ which indicates that we are 90% confident that the true population proportion $p$ is contained by the interval $(0.212, 0.295).$

2. The critical value for $\alpha = 0.05$ is $z_c = z_{1-\alpha/2} = 1.96.$

The corresponding confidence interval is computed as shown below:

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

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

$$=(\dfrac{76}{300}-1.96\sqrt{\dfrac{\dfrac{76}{300}(1-\dfrac{76}{300})}{300}},$$

$$\dfrac{76}{300}+1.96\sqrt{\dfrac{\dfrac{76}{300}(1-\dfrac{76}{300})}{300}})$$

$$=(0.204,0.303)$$

Therefore, based on the data provided, the 95% confidence interval for the population proportion is $0.204 < p < 0.303,$ which indicates that we are 95% confident that the true population proportion $p$ is contained by the interval $(0.204, 0.303).$

3. 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:

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

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

$$=(\dfrac{76}{300}-2.5758\sqrt{\dfrac{\dfrac{76}{300}(1-\dfrac{76}{300})}{300}},$$

$$\dfrac{76}{300}+2.5758\sqrt{\dfrac{\dfrac{76}{300}(1-\dfrac{76}{300})}{300}})$$

$$=(0.189,0.318)$$

Therefore, based on the data provided, the 99% confidence interval for the population proportion is $0.189< p < 0.318,$ which indicates that we are 99% confident that the true population proportion $p$ is contained by the interval $(0.189, 0.318).$