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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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