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A manufacturer of chocolate chips would like to know whether it’s bag filling machine works correctly at the 436 gram setting. It is believed that the machine is underfilling the bags. A 28 bag sample had a mean of 433 grams with a standard deviation of 23. A level of significance of 0.05 will be used. Assume the population distribution is approximately normal. Is there sufficient evidence to support the claim that the bags are under filled?
There is not sufficient evidence to support the claim that the bags are under filled.
Or
There is sufficient evidence to support the claim that the bags are under filled.

Respuesta :

Answer:

There is not sufficient evidence to support the claim that the bags are under filled.

Explanation:

We will perform a statistic evaluation to know if the filling machine is underfilling the bags.

Null hypothesis (H0): μ = 436 gram

Alternative hypothesis (Ha): μ < 436 gram

Because we do not know the population standard deviation, only the sample one, we will use the t-student test.

The decision rule is only for one tail because the alternative hypothesis is (<):

t-statistic< T(t-student table) --> You must accept the null hypothesis

t-statistic > T (t-student table) --> You must reject the null hypothesis

c) t-statistic formula:

t= (ybar-m)/(S/(sqrt(n)))

ybar: sample mean

m: hypothesized value

S: sample standard deviation

n: number of observations

t=(433-436)/(23/sqrt(28))

t= -0.69

The t-student table statistic at 5% significance level and for a sample of approx. 30 observations is: 1.6973. Because this test only uses the lower tail the statistic value is: -1.6973

-0.69<-1.6973

Then, you must accept the null hypothesis and reject the alternative hypothesis: there is not sufficient statistic evidence to support the claim that the bags are under filled.

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