Question

A study found that out of 300 people 60% of them prefer to eat hamburgers rather than hot dogs. Find the 95% confidence interval for the true proportion of people who prefer to eat hamburgers rather than hot dogs in the entire population.

Answer 1

(0.5446, 0.6554)

Explanation:

As the sample is sufficiently large, the formula is used to estimate the proportion shown in the attached image.

Where:

P: sample proportion = 0.6

n: Sample size = 300

`1-\alpha`: Confidence level = 0.95

α: Significance = 0.05

`Z_(\alpha / 2) = 1.96` *

* Obtained from the normal standard table.

When introducing these values in the formula shown in the image we obtain:

`0.6 + 1.96 *\sqrt {(0.6*0.4)/(300)}`

`0.6 - 1.96 *\sqrt{(0.6*0.4)/(300)}`

Finally, the confidence interval is:

(0.5446, 0.6554)