Answer:
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%.
Step-by-step 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%.
