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 words per minute with a standard deviation of 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 students was approximately 62.57 and 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

Set up the one-sample test

- H0: μ = 60 (new-method mean = old mean)
- H1: μ > 60 (new method increases speed)

Use a t-test (new-method population sd unknown). Sample: n = 30, x̄ = 62.57, s = 11.45.

Test statistic:
t = (x̄ − μ0) / (s/√n) = (62.57 − 60) / (11.45/√30)
√30 ≈ 5.477, s/√n ≈ 11.45/5.477 ≈ 2.091
t ≈ 2.57 / 2.091 ≈ 1.23 (df = 29)

One-sided p-value = P(T29 > 1.23) ≈ 0.11.

Because p ≈ 0.11 > 0.05, we fail to reject H0 at the 5% level. A 95% CI for μ is 62.57 ± t0.975,29·(s/√n) ≈ 62.57 ± 4.28 → (58.30, 66.85), which includes 60.

Conclusion: There is not enough evidence at the 5% level to conclude the new method increases words per minute. (The sample mean is higher, but variability is large, so the result is not statistically significant. A larger sample or reduced variability would be needed to detect a small effect reliably.)