Question

A poll was conducted in 2018 for people of different ages to see how many went hiking in the last year. The population mean in the United States of people ages 18 to 29 who went hiking in the past 12 months was 16.95. To figure out if this was true, I asked 10 randomly selected friends ages 18 to 29. Listed is how many times my friends went hiking in the past year: 2, 4, 1, 1, 5, 3, 2, 2, 0, 8. Is the mean of my friends less than the 2018 poll conducted? Use α = 0.05 level of significant.

what is the answer?

Answer 1

Data:    

n = 10   

μ = 16.95   

s = 2.3476   

x-bar = 2.8   

Hypotheses:    

Ho: μ ≥ 16.95   

Ha: μ < 16.95   

Decision Rule:    

α = 0.05   

Degrees of freedom = 10 - 1 = 9

Critical t- score = -1.833112923   

Reject Ho if t < -1.833112923   

Test Statistic:    

SE = s/√n = 2.3476/√10 = 0.742376304

t = (x-bar - μ)/SE = (2.8 - 16.95)/0.742376303501129 = -19.06041442

p- value = 0   

Decision (in terms of the hypotheses):   

There is sufficient evidence that the mean is less than 16.95

Since -19.06041442 < -1.833112923 we reject Ho and accept Ha

Conclusion (in terms of the problem):