A sample of 16 has mean 51.5 and standard deviation 6. 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?

Do not reject H₀: the data fit the claim

Reject H₀: the mean differs from the claim

Reject H₀ because t is positive

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

Question 66

HardOriginal question

A sample of 16 has mean 51.5 and standard deviation 6. 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

: the mean differs from 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 answer

Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

$t = \frac{51.5 - 50}{6/\sqrt{16}} = \frac{1.5}{1.5} = 1$ .
$\lvert t \rvert = 1 < 2.131$ : the difference could easily be chance, so $H_0$ is not rejected.

Worth noting

A sample mean different from 50 is expected by chance; only a large enough $t$ counts against $H_0$ .

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