Question

Irena Blazes, the fire chief of Flares, Oregon, wants to introduce volunteers into the department but fears that employees will resist volunteers. Chief Blazes would like to use volunteers and wants to avoid potential labor problems. She has decided to introduce volunteers if 60 percent of the employees agree to work with them. She administers a survey to all 150 paid firefighters, and 90 respond. Among these 90 firefighters, 62 say that they agree to work with volunteers and accept them in the work place. Given the results of this survey, should Chief Blazes procced with her plans to introduce volunteer firefighters? What is the probability that she could obtain the survey results if 60% or fewer of the paid firefighters were willing to accept volunteers?

Answer 1

Let p be the true proportion of paid firefighters were willing to accept volunteers.

Null Hypothesis H0: p = 0.6

Alternate hypothesis H1: p > 0.6

Given, Population Size, N = 150

Number of response, n = 90

Sample proportion, = 62 / 90 = 0.6889

Standard error of proportion for finite size of population is,

= 0.03277

Test Statistic, z = ( - p) / SE

= (0.6889 - 0.6) / 0.03277

= 2.71

P-value = P(z > 2.71) = 0.0034

Since p-value is less than 0.05 significance level, we reject null hypothesis H0 and conclude that there is significant evidence that true proportion of paid firefighters were willing to accept volunteers is greater than 0.6 (p > 0.6). Thus, Chief Blazes procced with her plans to introduce volunteer firefighters.

Probability that she could obtain the survey results if 60% or fewer of the paid firefighters were willing to accept volunteers is

P-value = 0.0034