Question

Previously, an organization reported that teenagers spent 24.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 25.7 hours with a sample standard deviation of 2.0. What is the appropriate test to perform?

Answer 1

We need to develop a one-tail t-student test ( test to the right )

We reject H₀ we find evidence that student spent more than 24,5 hours on the phone

Explanation:

Sample size n = 15 n < 30

And we were asked if the mean is higher than, therefore is a one-tail t-student test ( test to the right )

Population mean μ₀ = 24,5

Sample mean μ = 25,7

Sample standard deviation s = 2

Hypothesis Test:

Null Hypothesis H₀ μ = μ₀

Alternative Hypothesis Hₐ μ > μ₀

t (c) = ?

We will define CI = 95 % then α = 5 % α = 0,05 α/2 = 0,025

n = 15 then degree of freedom df = 14

From t-student table we get: t(c) = 2,1448

And t(s)

t(s) = ( μ - μ₀ ) / s/√n

t(s) = (25,7 - 24,5) /2/√15

t(s) = 2,3237

Now we compare t(c) and t(s)

t(c) = 2,1448 t(s) = 2,3237

t(s) > t(c)

Then we are in the rejection region we reject H₀ we have evidence at 95% of CI that students spend more than 24,5 hours per week on the phone