Answer to Question #169493 in Statistics and Probability for Sunera

Question

A random sample of 722 residents in a major town was asked whether they had ever been bitten by a dog. The responses (1=Yes and 2=N0) are recorded. Estimate with 95% confidence the proportion of residents who have been bitten by a dog.

Sample frequencies: n (1) = 304; n (2) = 418

Answer 1

The sample proportion is computed as follows, based on the sample size $N=722$  and the number of favorable cases $X=304$

$$\hat{p}=\dfrac{X}{N}=\dfrac{304}{722}\approx 0.42105$$

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:

$$CI(Proportion)$$

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

$$=(0.42105-1.96\sqrt{\dfrac{0.42105(1-0.42105)}{722}},$$

$$0.42105+1.96\sqrt{\dfrac{0.42105(1-0.42105)}{722}})$$

$$=(0.3850, 0.4571)$$

Therefore, based on the data provided, the 95% confidence interval for the population proportion is $0.3850<p<0.4571,$ which indicates that we are 95% confident that the true population proportion $p$ is contained by the interval $(0.3850, 0.4571).$