Question

Gloucester Township Department of Public Works, in an effort to encourage use of its public parks during the summer months, began a campaign to encourage vacationers to use the parks. Studies for the past five years indicate that only one-fourth of all vacationers use the park for three days or more per week. A survey conducted after the new campaign was initiated reveals that 55 of 200 vacationers use the parks three or more days per week. At α = .05, does this support the campaign’s effectiveness?

Answer 1

H₀: P = 0.25

H₁: P > 0.25

= 55/200 = 0.275

The test statistic z = ( - P)/sqrt(P(1 - P)/n)

                             = (0.275 - 0.25)/sqrt(0.25 * (1 - 0.25)/200)

                             = 0.82

P-value = P(Z > 0.82)

             = 1 - P(Z < 0.82)

             = 1 - 0.7939

             = 0.2061

Since the P-value is greater than the significance level(0.2061 > 0.05), so we should not reject H₀.

So at 5% significance level there is sufficient evidence to conclude that this support the campaign's effectiveness.