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Bottles of a popular cola are supposed to contain 300ml of cola. There is some variation from to bottle to bottle because the filling machinery is not perfectly precise. The distribution of the contents is normal with standard deviation of Q=3 ml. An inspector who suspects that the bottler is underfilling measures the contents of six bottles. The results are

299.4 297.7 301.0
298.9 300.2 297.0
Is this convincing evidence that the mean contents of cola bottles is less than the advertised 300 ml? perform a significance test at the Q= .05level.
Z=(300-299.03)/3/6= .79

1 Answer

1 vote

Answer: We accept H₀

There is not convincing evidence that the mean contents of cola is less thanthe advertised

Explanation:

Our case normal distribution, with σ = 3

The sample is of 6 then n = 6

The calculated mean from given data is : 299.03

sgnificanc level: α = 0.05

Then H₀ null hypothesis μ₀ = 300

the alternative hypothesis μₐ < 300

As n < 30 we have to go to t table we look fr α = 0.05 and gl = 5

we find t(c) = 2.015

Now we calculate t (s) = ( x - μ₀ )/( σ/√n ) t(e) = -0.97/1.224

t(s) = - 0.79

The test is one tail test ( to the left) and t(s) = -0,79 is bigger than

t (c) = -1.64

So we are in the acceptance zone. We accept H₀

answered
User Bhargav Thanki
by
8.4k points
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