Question

The Milky Cow farm had the average daily milk production per cow in 2016 equal to 60 liters. In 2017 the farm management decided to apply a diet resulting in the following production for 20 selected cows.

55.40	54.40	57.34	60.09	50.89	55.68	59.72	62.99	59.93	58.08
61.04	62.04	58.11	56.27	64.02	52.79	54.38	57.61	56.71	55.22

Setup the null and alternative hypotheses to test whether or not the mean of the milk volume has decreased.

Complete your test at the significance level α = 5%.               

What is the value of the test statistic (the standardized measure of the difference)?   

What is the critical value of the test statistic?

What is the p-value of the test.

Make the decision.

Answer 1

= (55.4 + 54.4 + 57.34 + 60.09 + 50.89 + 55.68 + 59.72 + 62.99 + 59.93 + 58.08 + 61.04 + 62.04 + 58.11 + 56.27 + 64.02 + 52.79 + 54.38 + 57.61 + 56.71 + 55.22)/20 = 57.6355

s = sqrt(((55.4 - 57.6355)^2 + (54.4 - 57.6355)^2 + (57.34 - 57.6355)^2 + (60.09 - 57.6355)^2 + (50.89 - 57.6355)^2 + (55.68 - 57.6355)^2 + (59.72 - 57.6355)^2 + (62.99 - 57.6355)^2 + (59.93 - 57.6355)^2 + (58.08 - 57.6355)^2 + (61.04 - 57.6355)^2 + (62.04 - 57.6355)^2 + (58.11 - 57.6355)^2 + (56.27 - 57.6355)^2 + (64.02 - 57.6355)^2 + (52.79 - 57.6355)^2 + (54.38 - 57.6355)^2 + (57.61 - 57.6355)^2 + (56.71 - 57.6355)^2 + (55.22 - 57.6355)^2)/19) = 3.4335

H₀: = 60

H₁: < 60

The test statistic t = ()/(s/)

                             = (57.6355 - 60)/(3.4335/)

                             = -3.08

At alpha = 0.05, the critical value is t_{0.05, 19} = -1.729

P-value = P(T < -3.08)

             = 0.0031

Since the test statistic value is less than the critical value (-3.08 < -1.729), so we should reject the null hypothesis.

So at 5% significance level there is sufficient evidence to conclude that the mean of the milk volume has decreased.