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A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 403 gram setting. It is believed that the machine is underfilling or overfilling the bags. A 10 bag sample had a mean of 411 grams with a variance of 121. Assume the population is normally distributed. Is there sufficient evidence at the 0.01 level that the bags are underfilled or overfilled

Sagot :

Answer:

There is not sufficient evidence at the 0.01 level that the bags are underfilled or overfilled

Step-by-step explanation:

A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 403 gram setting.

This means that the null hypothesis is:

[tex]H_{0}: \mu = 403[/tex]

Is there sufficient evidence at the 0.01 level that the bags are underfilled or overfilled:

This means that at the alternate hypothesis, we are testing if the mean is different than 403, that is:

[tex]H_{a}: \mu \neq 403[/tex]

The test statistic is:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

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

Value of 403 tested at the null hypothesis:

This means that [tex]\mu = 403[/tex]

A 10 bag sample had a mean of 411 grams with a variance of 121.

This means that [tex]n = 10, X = 411, \sigma = \sqrt{121} = 11[/tex]

Value of the test statistic:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]z = \frac{411 - 403}{\frac{11}{\sqrt{10}}}[/tex]

[tex]z = 2.3[/tex]

Pvalue:

Testing if the mean is different of a value, and z positive, which means that the pvalue is 2 multiplied by 1 subtracted by the pvalue of z = 2.3

Looking at the z-table, z = 2.3 has a pvalue of 0.9893

1 - 0.9893 = 0.0107

2*0.0107 = 0.0214

0.0214 > 0.01, which means that there is not sufficient evidence at the 0.01 level that the bags are underfilled or overfilled