A sample of 9 packets has mean weight 106 g and standard deviation 6 g. The maker claims a mean of 100 g. Testing H₀: μ = 100 against H₁: μ ≠ 100 at the 5% level, with critical value t = 2.306 for 8 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 65

HardOriginal question

A sample of 9 packets has mean weight 106 g and standard deviation 6 g. The maker claims a mean of 100 g. Testing $H_0: \mu = 100$ against $H_1: \mu \ne 100$ at the 5% level, with critical value $t = 2.306$ for 8 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

$t = \frac{106 - 100}{6/\sqrt{9}} = \frac{6}{2} = 3$ .
$\lvert t \rvert = 3 > 2.306$ , so the difference is significant: reject $H_0$ .

Worth noting

The decision rests on the size of $t$ against 2.306, not on its sign.

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