Question

conduct the appropriate hypothesis test and compute the p-value.

a media research company claims that 85% of gen-z only acquires news from the internet, rather than television, radio, or print. a broadcaster believes this estimate to be high and collects data on a random sample of 500 individuals from gen-z. of the 500 people, 410 said they only acquire news from the internet.

at the 0.05 significance level, is there strong enough sample evidence to suggest that the proportion is less than 0.85?

yes, because the p-value = 0.04
yes, because the p-value = 0.03
no, because the p-value = 0.03
no, because the p-value = 0.04

Explanation:

Step1: Identify Hypotheses

Null hypothesis \( H_0: p = 0.85 \), Alternative hypothesis \( H_1: p < 0.85 \) (left - tailed test). Sample proportion \( \hat{p}=\frac{410}{500} = 0.82 \), sample size \( n = 500 \), hypothesized proportion \( p_0=0.85 \).

Step2: Calculate Test Statistic

The formula for the z - statistic in a proportion test is \( z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \)

Substitute values: \( \hat{p} = 0.82 \), \( p_0=0.85 \), \( n = 500 \)

First, calculate the denominator: \( \sqrt{\frac{0.85\times(1 - 0.85)}{500}}=\sqrt{\frac{0.85\times0.15}{500}}=\sqrt{\frac{0.1275}{500}}=\sqrt{0.000255}\approx0.01597 \)

Then, the numerator: \( 0.82-0.85=- 0.03 \)

So, \( z=\frac{-0.03}{0.01597}\approx - 1.88 \)

Step3: Find p - value

For a left - tailed test, the p - value is \( P(Z < z) \), where \( z=-1.88 \)

Using the standard normal table, \( P(Z < - 1.88)=1 - P(Z < 1.88) \)

From the standard normal table, \( P(Z < 1.88) = 0.9699 \), so \( P(Z < - 1.88)=1 - 0.9699 = 0.0301\approx0.03 \)

Since the p - value (\( 0.03 \)) is less than the significance level \( \alpha = 0.05 \), we reject the null hypothesis. So there is enough evidence to suggest that the proportion is less than 0.85.

Answer:

Yes, because the p - value = 0.03