Question

It is believed that 30% of the people in Washington state want cell phone use banned in cafes. The CEO of a major coffehouse chain in Seattle wonders whether the opinion of the people who go to her cafes is different from the overall population. She hires a polling agency to investigate this issue. In a random sample of 1450 individuals, 474 people have the same opinion as the overall population. What is the p-value?

Answer 1

The p-value is 0.025

Explanation:

Given that:

30% of the people in Washington state want cell phone use banned in cafes

sample size N = 1450

X = 474 people have the same opinion.

The objective is to calculate the p-value.

Let's assume the level of significance = 5% = 0.05

From the information given ; the null and alternative hypothesis can be computed as:

`\mathbf{H_o: p = 0.30 } \\ \\ \mathbf{H_a: p \\eq 0.30}}`

USING MINITAB; the simulation on what we compute on our MINITAB can be written as:

Test for p = 0.3 vs p not = 0.3

Sample X N Sample p 95 C.I Z-value P-value

1 474 1450 0.326897 (0.302752, 0.351041) 2.23 0.025

From what we have in our MINITAB Output;

Z - value = 2.23

P-Value = 0.025