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.

yo answer should be 6 if im not mistaken

step-by-step explanation:

answer: 1,380

step-by-step explanation:

1: you turn %15 into a decimal by moving the decimal point two places to the left.

2: and then you multiply .15 by 1,200 , which will get you 180.00

3: from there you will add 180 to 1,200

4: which will get you 1,380

hope it *im kind of slow so idk if my answer is right*