Question

It is claimed that the proportion of people who favor the use of oxygenated fuels year around to reduce air pollution is 60%. A pollster believes this proportion to be actually higher than 60%. She polls a random sample of 500 registered voters and finds out that 315 of them favor the use of oxygenated fuels. At level of significance 0.09, what is her conclusion?

Answer 1

Conclusion is that there is sufficient evidence to reject the claim that the proportion of people who favor the use of oxygenated fuels year around to reduce air pollution is 60%.

Explanation:

We are given;

Population proportion; p = 60% = 0.6

Sample mean: x¯ = 315

Sample size; n = 500

Significance level = 0.09

Defining the hypothesis, we have;

Null Hypothesis; H0: p = 0.6

Alternative hypothesis; Ha: p > 0.6

z-score formula in this case is;

z = (x¯ - np)/√(np(1 - p))

Plugging in the relevant values;

z = (315 - (500 × 0.6))/√(500 × 0.6(1 - 0.6))

z = 1.72

From z-distribution table, the p-value at z = 1.72 is;

1 - P(Z ≤ z-score)

This gives:

p-value = 1 - 0.95728 = 0.04272

Thus p-value is less than the significance level of 0.09, thus we will reject the null hypothesis and conclude that there is sufficient evidence to reject the claim that the proportion of people who favor the use of oxygenated fuels year around to reduce air pollution is 60%.