A sample of 16 batteries lasts 46 hours on average, with standard deviation 4 hours. The maker claims 50 hours. Testing H₀: μ = 50 against H₁: μ ≠ 50 at the 5% level, with critical value t = 2.131 for 15 degrees of freedom, what is the decision?

Reject H₀: the mean differs from the claim

Do not reject H₀: the data fit the claim

Reject H₀ because t is positive

Accept H₁ because the sample mean is not equal to the claim

Question 68

HardOriginal question

A sample of 16 batteries lasts 46 hours on average, with standard deviation 4 hours. The maker claims 50 hours. Testing $H_0: \mu = 50$ against $H_1: \mu \ne 50$ at the 5% level, with critical value $t = 2.131$ for 15 degrees of freedom, what is the decision?

Do not reject

H_0

: the data fit the claimCorrect answerYour answerReject

H_0

because

t

is positiveCorrect answerYour answerAccept

H_1

because the sample mean is not equal to the claimCorrect answerYour answerReject

H_0

: the mean differs from the claimCorrect answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer D

Correct answer: D

$t = \frac{46 - 50}{4/\sqrt{16}} = \frac{-4}{1} = -4$ .
$\lvert t \rvert = 4 > 2.131$ : reject $H_0$ ; the batteries last less than claimed.

Worth noting

A negative $t$ is compared by its size, $\lvert t \rvert$ , in a two-tailed test.

Ten questions from Sampling, Inference & the t-Test, at your own pace.