Answer to Question #253176 in Statistics and Probability for hya

Question

A nutritionist wants to know the population of Kindergarten pupils who eat

vegetables. A survey among 1500 kinder pupils was conducted and revealed

that 485 pupils ate vegetables. Construct the 99% confidence interval to

determine the true proportion of pupils who eat vegetables. Interpret the data.

Answer 1

The sample proportion is computed as follows, based on the sample size $N = 1500$ and the number of favorable cases $X = 485:$

$$\hat{p}=\dfrac{485}{1500}=\dfrac{97}{300}\approx0.3233$$

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:

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

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

$$=(\dfrac{97}{300}-2.5758\sqrt{\dfrac{\dfrac{97}{300}(1-\dfrac{97}{300})}{1500}},$$

$$\dfrac{97}{300}+2.5758\sqrt{\dfrac{\dfrac{97}{300}(1-\dfrac{97}{300})}{1500}})$$

$$=(0.2922,0.3544)$$

Therefore, based on the data provided, the 99% confidence interval for the population proportion is $0.2922 < p < 0.3544,$ which indicates that we are 99% confident that the true population proportion $p$  is contained by the interval $(0.2922, 0.3544).$