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g A sanitation supervisor is interested in testing to see if the mean amount of garbage per bin is different from 50. In a random sample of 36 bins, the sample mean amount was 50.67 pounds and the sample standard deviation was 3.9 pounds. Conduct the appropriate hypothesis test using a 0.01 level of significance.

Sagot :

Answer:

We accept the null hypothesis, that is, the the mean amount of garbage per bin is not different from 50.

Step-by-step explanation:

A sanitation supervisor is interested in testing to see if the mean amount of garbage per bin is different from 50.

This means that the null hypothesis is:

[tex]H_{0}: \mu = 50[/tex]

And the alternate hypothesis is:

[tex]H_{a}: \mu \neq 50[/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.

50 tested at the null hypothesis:

This means that [tex]\mu = 50[/tex]

In a random sample of 36 bins, the sample mean amount was 50.67 pounds and the sample standard deviation was 3.9 pounds.

This means that [tex]n = 36, X = 50.67, \sigma = 3.9[/tex]

Value of the test statistic:

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

[tex]z = \frac{50.67 - 50}{\frac{3.9}{\sqrt{36}}}[/tex]

[tex]z = 1.03[/tex]

p-value:

Since we are testing if the mean is differente from a value and the z-score is positive, the pvalue is 2 multipled by 1 subtracted by the pvalue of z = 1.03.

z = 1.03 has a pvalue of 0.8485

1 - 0.8485 = 0.1515

2*0.1515 = 0.3030

0.3030 > 0.01, which means that we accept the null hypothesis, that is, the the mean amount of garbage per bin is not different from 50.

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