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

Answer:

(0.5446, 0.6554)

Step-by-step 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:

$0.6 + 1.96 *\sqrt {\frac{0.6*0.4}{300}}$

$0.6 - 1.96 *\sqrt{\frac{0.6*0.4}{300}}$

Finally, the confidence interval is:

(0.5446, 0.6554)