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 48.99 pounds and the sample standard deviation was 3.7 pounds. Conduct the appropriate hypothesis test using a 0.01 level of significance.

a) What is the test statistic? Give your answer to four decimal places.

b) What is the P-value for the test? Give your answer to four decimal places.

Step-by-step explanation:

Claim: if the mean amount of garbage per bin is different from 50.

Null hypothesis: u=50

Alternative hypothesis : u = / 50

Using the z score formular for a one sample z test - z = (x - u) / (sd/√n)

Where x = 48.99, u = 50 sd = 3.7 and n = 36

z = 48.99 - 50 / (3.7/√36)

z = - 1.01 / (3.7/6)

z = - 1.01/0.6167

z = - 1.6377

To find the p value at a 0.01 level of significant from the - 1.6377 z score for a two tailed test the p value using the p value calculator is 0.1016. The result is not significant at 0.01 level of significant thus we will fail to reject the null and conclude that the mean amount of garbage per bin is 50.