Question

A federal agency responsible for enforcing laws governing weights and measures routinely inspects package to determine whether the weight of the contents is at least as great as that advertised on the package. A random sample of 18 containers whose packing states that the contents weigh 8 ounces was drawn. The contents were weighted and the results follow.: sample mean=7.91, and sample standard deviation=0.085. Can we conclude at 1% significance level that on average the containers are mislabeled?

Answer 1

Answer:

The calculated value |t|  = 4.493>2.567 at 0.01 level of significance with

17 degrees of freedom.

Therefore the null hypothesis is rejected.

The weight of the contents is not at least as great as that advertised on the package

Step-by-step explanation:

Step:-(i)

Given data a random sample of 18 containers whose packing states that the contents weigh 8 ounces was drawn.

sample size 'n' = 18

mean of the population 'μ' = 8

sample mean (x⁻) =7.91, and

sample standard deviation (S) = 0.085.

Step:-(ii)

Given data the weight of the contents is at least as great as that advertised on the package That is

Null hypothesis : 'μ' > 8

Alternative hypothesis:μ' < 8

Level of significance α=0.01

Test statistic :-

$t = \frac{x^{-} -u}{\frac{S}{\sqrt{n} } }$

$t = \frac{7.91 -8}{\frac{0.085}{\sqrt{18} } }$

t = -4.493

|t| =|-4.493| = 4.493

Degrees of freedom of t- distribution γ =n-1 = 18-1=17

The tabulated value t= 2.567 at 0.01 level of significance.

Conclusion:-

The calculated value |t|  = 4.493>2.567 at 0.01 level of significance with

17 degrees of freedom.

Therefore the null hypothesis is rejected.

The alternative hypothesis is accepted

The weight of the contents is not at least as great as that advertised on the package.