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A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 444 gram setting. It is believed that the machine is underfilling the bags. A 9 bag sample had a mean of 441 grams with a standard deviation of 25. A level of significance of 0.025 will be used. Assume the population distribution is approximately normal.

Required:
Is there sufficient evidence to support the claim that the bags are underfilled?

Sagot :

Answer:

The pvalue of the test is 0.3641 > 0.025, which means that there is not sufficient evidence to support the claim that the bags are underfilled.

Step-by-step explanation:

A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 444 gram setting. Test if they are underfilled:

At the null hypothesis, we test if the mean is the correct mean of 444, that is:

[tex]H_0: \mu = 444[/tex]

At the alternate hypothesis, we test if it is underfilled, that is, the mean is less than 444.

[tex]H_a: \mu < 444[/tex]

The test statistic is:

[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]

In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

444 is tested at the null hypothesis:

This means that [tex]\mu = 444[/tex]

A 9 bag sample had a mean of 441 grams with a standard deviation of 25.

This means that [tex]n = 9, X = 441, s = 25[/tex]

Value of the test statistic:

[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]

[tex]t = \frac{441 - 444}{\frac{25}{\sqrt{9}}}[/tex]

[tex]t = -0.36[/tex]

Pvalue of the test and decision:

The pvalue of the test is the probability of finding a mean of 441 or less, which is the pvalue of t = -0.36, with 9 - 1 = 8 degrees of freedom on a one-tailed test.

Using a calculator, the pvalue is 0.3641.

The pvalue of the test is 0.3641 > 0.025, which means that there is not sufficient evidence to support the claim that the bags are underfilled.