Question

17) The Test Environment for Optimal Performance (TEOP) is a measure that assesses the students preferred testing conditions, namely sounds and actions. Scores range from 1 to 5, with 5 being the preference for sound or action. The mean score for U.S. college students sound preference is 2.59 and the standard deviation is .605. A teacher who suspects that older students have lower preferences toward sounds gives the TEOP to 351 high school students. The mean score for the high school sample was 2.75.

• State the null and alternative hypotheses:
• Calculate the z-statistic (Z_o):
• Should you reject or fail to reject the null hypothesis and why if your level of significance is .05 (α = .05)?
• Why?
• What does this mean?

Answer 1

The null and alternative hypotheses:

H0: μ = 2.59

H1: μ > 2.59

The z-statistic (Zo) = (x̅ - μ)/(σ/√n)

Substituting x̅ = 2.75, μ = 2.59, σ = 0.605 and n = 351

The z-statistic (Zo) = (2.75-2.59)/(0.605/sqrt(351))

The z-statistic (Zo) = 4.9547

P-value correspodning z = 4.9547 for a right tailed test is obtained using calculator. Screenshot below:

P-value = 0.0000

Since p-value = 0.0000 < α = 0.05, we reject the null hypothesis

At 0.05 significance level, there is enough evidence to conclude that older students have lower preferences toward sounds.