Question

In a poll of 1000 adults in july​ 2010, 530 of those polled said that schools should ban sugary snacks and soft drinks. complete parts a and b below.

a. do a majority of adults​ (more than​ 50%) support a ban on sugary snacks and soft​ drinks? perform a hypothesis test using a significance level of 0.05. state the null and alternative hypotheses. note that p is defined as the population proportion of people who believe that schools should ban sugary foods.

Answer 1

Given:

n = 1000

Sample proportion, p' = 530/1000 = 0.53

P = 0.5

significance level, a= 0..05

For null hypothesis and alternative hypothesis:

`H_0 : p=0.5`

`H_a : p > 0.5`

For test statistic, Z, we have:

`\frac{p' - p}{\sqrt{(p*(1-p))/(n)}}`

`\frac{0.53 - 0.5}{\sqrt{(0.5(1-0.5))/(1000)}}`

Zscore = 1.898 ≈ 1.90

p-value = P(Z > 1.9) = 1 -P(<1.9)

= 1 - 0.9713 = 0.0287

Therefore, p- value = 0.0287

Since the p- value(0.0287) is less than significance level (0.05), we reject null hypothesis, H0.