Question

A random sample of 200 tins of coconut oil gave an average weight of 4.95kgs with a standard deviation of 0.21kg. Do we accept the hypothesis of net weight 5kgs per tin at 1% level?

Answer 1

Step-by-step explanation:

A random sample of 200 tins of coconut oil gave an average weight of 4.95kgs with S.D of 0.21kg. Do we accept the hypothesis of net weight 5 kgs ...

Answer 2

Answer:

We cannot accept the hypothesis of net weight is 5kg per tin at 1% level.

Step-by-step explanation:

Given the size of the random sample, n = 200

Average weight,  $\overline{x}= 4.95kg$

Standard deviation, $\sigma =0.21kg$

The net weight,  $\mu=5kgs$

The null hypothesis $H_o$ is that the net weight of each tin is 5 kg.

A z - test can be used to test an alternated hypothesis against the stated null hypothesis.

The z-test formula is given

$z=\dfrac{\bar x - \mu}{\sigma/\sqrt{n} }$

Therefore, let the null hypothesis, $H_o : \mu = 5 kg$

and alternate hypothesis,  $H_1 : \mu\ne 5 kg$

Substituting the values in z-test formula,

$z=\dfrac{4.95 - 5}{0.21/\sqrt{200} }$

  $=\dfrac{-0.0 5}{0.21}\times 10\sqrt{2}$

  $=-2.38\times \sqrt{2}\approx-3.37$

For 1% level of significance:

$|z|=3.37 > 2.58$

Hence, the null hypothesis $H_o$ is rejected.

Therefore, we cannot consider the net weight is 5 kg per tin at 1% level of significance.

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