Question

answer in the simplest way and like a teen would understand
A manufacturer claims that their battery for a particular cell phone model ​lasts
hours.
​
​However, consumers complained that the batteries lasted ​less than
hours.
​
​The manufacturer randomly sampled
batteries and ​found that the mean battery life of the sample was
hours with a
​standard deviation of
hours.
​
​Based on the data given, is there evidence ​that the manufacturer’s claim is false?
​Justify your answer.

Answer 1

I can’t finish the test without the actual numbers (the claimed hours, the sample size n, the sample mean, and the sample standard deviation). Give me those and I’ll compute it.

Meanwhile, here’s the simplest way to do it — like a teen would understand:

1. Figure out the claim and the complaint:
- Manufacturer claims the mean battery life = μ0 hours.
- Consumers say it’s less, so we test: H0: μ = μ0 (or μ ≥ μ0) vs H1: μ < μ0.

2. Use a t-test (because we use the sample standard deviation):
- Compute t = (sample mean − μ0) / (s / √n)
- sample mean = x̄, s = sample standard deviation, n = sample size.

3. Find the p-value:
- p-value = probability that a t random variable with (n − 1) degrees of freedom is ≤ your t.
- (You can get this from a calculator, Excel, or an online t-table.)

4. Decide:
- Pick a significance level (usually 0.05).
- If p-value < 0.05, there’s evidence the batteries last less than the claimed hours → manufacturer’s claim is likely false.
- If p-value ≥ 0.05, there’s not enough evidence → we don’t reject the claim.

Quick example (so you see how it works):
- Suppose claimed μ0 = 10 hours, n = 16, x̄ = 9.2 hours, s = 1.1 hours.
- t = (9.2 − 10) / (1.1 / √16) = (−0.8) / (0.275) ≈ −2.91.
- With df = 15, p-value ≈ 0.005 (one-tailed).
- p < 0.05 → evidence batteries last less than 10 hours → claim likely false.

Give me the real numbers from your problem and I’ll do the exact calculation and conclusion.