Asked by c
answer in the simplest way and like a teen would understand
A manufacturer claims that their battery for a particular cell phone model lasts 10 hours.
However, consumers complained that the batteries lasted less than 10 hours.
The manufacturer randomly sampled 24 batteries and found that the mean battery life of the sample was 9.4 hours with astandard deviation of 0.25 hours.
Based on the data given, is there evidence that the manufacturer’s claim is false?
Justify your answer.
can you actually put the math right instead of just slahes like a actual fraction
All Answers 1
Answered by
GPT-5 mini
AI
Short answer: Yes — strong evidence the batteries last less than 10 hours.
Work shown (simple math, using fractions):
1) Hypotheses
- H0: μ = 10 (manufacturer is right)
- Ha: μ < 10 (batteries last less than 10 hours)
2) Test statistic (one‑sample t)
t = (x̄ − μ0) / (s / √n)
= (9.4 − 10) / (0.25 / √24)
Write the numbers as fractions:
9.4 − 10 = −0.6 = −3/5, and 0.25 = 1/4, so
denominator = (1/4) / √24 = 1 / (4√24).
So
t = (−3/5) ÷ (1 / (4√24)) = (−3/5) × (4√24) = (−12√24) / 5.
Now √24 = 2√6 ≈ 4.89898, so
t ≈ (−12 × 4.89898) / 5 ≈ −11.76.
Degrees of freedom = n − 1 = 23.
3) P‑value and conclusion
A t of −11.76 with 23 df gives a p‑value essentially 0 (way less than 0.001). That means the result is extremely unlikely if the true mean were 10. So we reject H0 and conclude there is strong evidence the batteries last less than 10 hours.
In teen words: the sample mean (9.4) is far below 10 given how small the sample spread was, so the data say the claim is false.
Work shown (simple math, using fractions):
1) Hypotheses
- H0: μ = 10 (manufacturer is right)
- Ha: μ < 10 (batteries last less than 10 hours)
2) Test statistic (one‑sample t)
t = (x̄ − μ0) / (s / √n)
= (9.4 − 10) / (0.25 / √24)
Write the numbers as fractions:
9.4 − 10 = −0.6 = −3/5, and 0.25 = 1/4, so
denominator = (1/4) / √24 = 1 / (4√24).
So
t = (−3/5) ÷ (1 / (4√24)) = (−3/5) × (4√24) = (−12√24) / 5.
Now √24 = 2√6 ≈ 4.89898, so
t ≈ (−12 × 4.89898) / 5 ≈ −11.76.
Degrees of freedom = n − 1 = 23.
3) P‑value and conclusion
A t of −11.76 with 23 df gives a p‑value essentially 0 (way less than 0.001). That means the result is extremely unlikely if the true mean were 10. So we reject H0 and conclude there is strong evidence the batteries last less than 10 hours.
In teen words: the sample mean (9.4) is far below 10 given how small the sample spread was, so the data say the claim is false.
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