Question

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 batteries and found that the mean battery life of the sample was 9.4 hours with a
standard deviation of 0.25 hours.

Based on the data given, is there evidence that the manufacturer’s claim is false?
Justify your answer.

Answer 1

You need one piece of missing information to get a numeric p-value: the sample size n (and a significance level α if you want a decision rule). With that said, here is the correct test and what the data imply.

1) Hypotheses
- H0: μ = 10 (manufacturer’s claim)
- Ha: μ < 10 (consumers claim battery life is less than 10 hours)

2) Test statistic (use t because σ is unknown)
t = (x̄ − μ0) / (s / √n) = (9.4 − 10) / (0.25 / √n) = −0.6 / (0.25/√n) = −2.4√n.

3) Conclusion depends on n
- If you tell me n I will compute the exact t-value, degrees of freedom (df = n − 1) and one-sided p-value.
- For any moderately sized sample the result is highly significant because the sample mean is 0.6 hours below the claimed mean while the sample standard deviation is only 0.25 hours.

Illustrative examples:
- If n = 3: t = −2.4√3 ≈ −4.16, df = 2, one-sided p ≈ 0.03 → reject H0 at α = 0.05.
- If n = 10: t = −2.4√10 ≈ −7.59, df = 9, p ≪ 0.001 → very strong evidence that μ < 10.
- Even for n = 2: t ≈ −3.39, df = 1, one-sided p ≈ 0.06 → not quite significant at 5% (but close).

So: unless the manufacturer’s “random sample” consisted of only 1 or 2 batteries, the observed data provide strong evidence that the true mean battery life is less than 10 hours. If you give the actual sample size (and a significance level if you want a formal decision), I will compute the exact p-value and decision.