Question

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.

Answer 1

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

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:

`H_(0): \mu = 50`

And the alternate hypothesis is:

`H_(a): \mu \\eq 50`

The test statistic is:

`z = (X - \mu)/((\sigma)/(√(n)))`

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

50 tested at the null hypothesis:

This means that
`\mu = 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.

This means that
`n = 36, X = 50.67, \sigma = 3.9`

Value of the test statistic:

`z = (X - \mu)/((\sigma)/(√(n)))`

`z = (50.67 - 50)/((3.9)/(√(36)))`

`z = 1.03`

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.