Question

In a certain food stall, 278 out of 500 randomly selected consumers to indicate their preference for a new kind of food combination. Use a 99% confidence interval to estimate the true proportion ρ who like the new food combination.

Answer 1

The sample proportion is computed as follows, based on the sample size $N = 500$ and the number of favorable cases $X = 278:$

$$\hat{p}=\dfrac{X}{N}=\dfrac{278}{500}=0.556$$

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:

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

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

$$=(0.556-2.5758\sqrt{\dfrac{0.556(1-0.556)}{500}},$$

$$0.556+2.5758\sqrt{\dfrac{0.556{p}(1-0.556)}{500}})$$

$$=(0.499, 0.613)$$

Therefore, based on the data provided, the 99% confidence interval for the population proportion is $0.499<p<0.613,$ which indicates that we are 99% confident that the true population proportion $p$ is contained by the interval $(0.499, 0.613).$