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answer in the most simplest way and like a teen and explain word fr word
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.
All Answers 1
Answered by
GPT-5 mini
AI
Alright, I’ll keep it super simple and explain step-by-step like a teen.
Goal: See if the new method makes students type faster than 60 wpm.
Set up (plain):
- Null idea (H0): average speed is 60 wpm (no improvement).
- What we want (HA): average speed is more than 60 wpm (improvement).
Given:
- Old method mean = 60 wpm.
- Old method std dev = 3.5 wpm (this is important — more on that below).
- New sample: n = 30, sample mean = 62.57 wpm, sample std dev = 11.45.
Way 1 — if we trust the old std dev (3.5) as the true population std dev:
1. Standard error = 3.5 / sqrt(30) ≈ 0.638.
2. z = (62.57 − 60) / 0.638 ≈ 4.03.
3. That z gives a one-sided p-value ≈ 0.00003 (super tiny).
4. So we reject H0 and say: yes, the new method is effective (it increases wpm).
Way 2 — if we do NOT trust the old std dev and use the sample std dev (11.45):
1. Standard error = 11.45 / sqrt(30) ≈ 2.09.
2. t = (62.57 − 60) / 2.09 ≈ 1.23 with df = 29.
3. One-sided p-value ≈ 0.11.
4. That’s not small enough, so we fail to reject H0 — we would NOT conclude the new method is effective.
Which is right?
- If the old std dev (3.5) really is the true population spread, use Way 1 → clear evidence the method works.
- But the new sample’s std dev (11.45) is much bigger, so it's suspicious to assume 3.5 still applies. If you’re careful and use the sample std dev, you don’t get strong evidence.
Short answer: If you assume the old standard deviation (3.5) is correct, then yes — the new method looks effective. If you don’t trust that and use the sample’s bigger spread, then no — the result isn’t convincing.
Goal: See if the new method makes students type faster than 60 wpm.
Set up (plain):
- Null idea (H0): average speed is 60 wpm (no improvement).
- What we want (HA): average speed is more than 60 wpm (improvement).
Given:
- Old method mean = 60 wpm.
- Old method std dev = 3.5 wpm (this is important — more on that below).
- New sample: n = 30, sample mean = 62.57 wpm, sample std dev = 11.45.
Way 1 — if we trust the old std dev (3.5) as the true population std dev:
1. Standard error = 3.5 / sqrt(30) ≈ 0.638.
2. z = (62.57 − 60) / 0.638 ≈ 4.03.
3. That z gives a one-sided p-value ≈ 0.00003 (super tiny).
4. So we reject H0 and say: yes, the new method is effective (it increases wpm).
Way 2 — if we do NOT trust the old std dev and use the sample std dev (11.45):
1. Standard error = 11.45 / sqrt(30) ≈ 2.09.
2. t = (62.57 − 60) / 2.09 ≈ 1.23 with df = 29.
3. One-sided p-value ≈ 0.11.
4. That’s not small enough, so we fail to reject H0 — we would NOT conclude the new method is effective.
Which is right?
- If the old std dev (3.5) really is the true population spread, use Way 1 → clear evidence the method works.
- But the new sample’s std dev (11.45) is much bigger, so it's suspicious to assume 3.5 still applies. If you’re careful and use the sample std dev, you don’t get strong evidence.
Short answer: If you assume the old standard deviation (3.5) is correct, then yes — the new method looks effective. If you don’t trust that and use the sample’s bigger spread, then no — the result isn’t convincing.
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