Answer to Question #350957 in Statistics and Probability for SimplyJhoyz

Question

A nutritionist wants to know the populatiin proportion of grade 11 students who eat vegetables. Pegged at a confidence of 95% survey among 1200 respondents was conducted and 200 said that they eat vegetables.

a.What is the margin of error?

b.Construct the confidence interval

c.Find the length of the confidence interval

Answer 1

a.

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

$$ME=z_c\times\sqrt{\dfrac{\hat{p}(1-\hat{p}}{n}}$$

$$\hat{p}=\dfrac{x}{n}=\dfrac{200}{1200}=\dfrac{1}{6}$$

$$ME=1.96\times\sqrt{\dfrac{\dfrac{1}{6}(1-\dfrac{1}{6})}{1200}}\approx0.021$$

b. The corresponding confidence interval is computed as shown below:

$$CI(Proportion)=(\hat{p}-ME, \hat{p}+ME)$$

$$=(\dfrac{1}{6}-0.021, \dfrac{1}{6}+0.021)$$

$$=(0.146, 0.188)$$

Therefore, based on the data provided, the 95% confidence interval for the population proportion is $0.146 < p < 0.188,$ which indicates that we are 95% confident that the true population proportion $p$ is contained by the interval $(0.146, 0.188).$

c. The length of the confidence interval is $2ME=0.042.$