Question

A keyboarding instructor wants to test if a new method of instruction is effective.



With the old method students typed a mean of
60
60 60 words per minute with a standard deviation of 3.5
3.5
3.5 words per minute.



Thirty students selected at random received instruction using the new method and were then given a typing test.



The mean and standard deviation of the typing speeds of the
30
30 students was approximately 62.57and
11.45
11.45, respectively.



Would you conclude the new method of instruction is effective in increasing the number of words per minute?



Explain your answer.

Answer 1

We test H0: μ = 60 (no change) vs Ha: μ > 60 (new method increases WPM). Use the sample of n = 30, x̄ = 62.57, s = 11.45.

Test statistic (one‑sample t):
SE = s/√n = 11.45/√30 ≈ 2.0905
t = (62.57 − 60)/SE ≈ 2.57/2.0905 ≈ 1.23
df = 29.

One‑sided p ≈ 0.11 (≈ 0.113). At α = 0.05 we fail to reject H0. There is not enough evidence to conclude the new method increases typing speed.

(As a check, a 95% CI for μ is 62.57 ± t0.975,29·SE ≈ 62.57 ± 4.28 → (58.29, 66.85), which contains 60.)

Note: If you incorrectly treat the old method’s σ = 3.5 as the true known population SD and use a z‑test, you would get a highly significant result (z ≈ 4.02), but given the new sample’s much larger observed SD (11.45) it is more appropriate to use the t‑test based on the sample SD.