Question

answer in the most simplest way and sound like a teen
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

Yep — looks like the new method works.

Why:
- Hypotheses: H0: μ = 60 (no change) vs Ha: μ > 60 (increase).
- They gave the old method’s standard deviation (σ = 3.5), so use a z-test.
- z = (62.57 − 60) / (3.5 / √30) = 2.57 / 0.639 ≈ 4.02.
- That z gives a one-sided p-value ≈ 0.00003, which is way less than 0.05.

Conclusion: reject H0 — the mean typing speed increased, so the new method is effective.

(If you instead treated σ as unknown and used the sample SD = 11.45 with a t-test, the result wouldn’t be significant. But using the given old-method σ = 3.5 is the usual approach here.)