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.

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:

And the alternate hypothesis is:

The test statistic is:

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

50 tested at the null hypothesis:

This means that

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

Value of the test statistic:

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.

is there any attachment? what are the equations?

answer: 2 and 1

step-by-step explanation: got this question