Question

Babcock and Marks (2010) reviewed survey data from 2003–2005 and obtained an average of μ = 14 hours per week spent studying by full-time students at four-year colleges in the United States. To determine whether this average has changed in the past 10 years, a researcher selected a sample of n = 64 of today’s college students and obtained an average of M = 12.5 hours. If the standard deviation for the distribution is σ = 4.8 hours per week, does this sample indicate a significant change in the number of hours spent studying?

Answer 1

We do not have enough evidence to accept H₀

Explanation:

Normal Distribution

size sample = n = 64 (very small sample for evaluating population of 5 years

Standard deviation 4,8

1.- Test hypothesis

H₀ null hypothesis ⇒ μ₀ = 14 and

Hₐ alternative hypothesis ⇒ μ₀ ≠ 14

2.- z(c) we assume α = 0,05 as we are dealing with a two test tail we should consider α/2 = 0.025.

From z table we the z(c) value

z(c) = 1.96 and of course by symmetry z(c) = -1.96

3.- We proceed to compute z(s)

z(s) = [ ( μ - μ₀ ) /( σ/√n) ] ⇒ z(s) = - (1.5)*√64/4.8

z(s) = - 2.5

We compare z(s) and z(c)

z(s) < z(c) -2.5 < -1.96 meaning z(s) is in the rejection zone

we reject H₀ .

From the start we indicate sample size as to small for the experiment nonetheless we found that we dont have enough evidence to accept H₀