Question

A farmer uses a lot of fertilizer to grow his crops. The farmer's manager thinks fertilizer products from distributor A contain more of the nitrogen that his plants need than distributor B's fertilizer does. He takes two independent samples of four batches of fertilizer from each distributor and measures the amount of nitrogen in each batch. Fertilizer from distributor A contained 23 pounds per batch and fertilizer from distributor B contained 18 pounds per batch. Suppose the population standard deviation for distributor A and distributor B is four pounds per batch and five pounds per batch, respectively. Assume the distribution of nitrogen in fertilizer is normally distributed. Let µ1 and µ2 represent the average amount of nitrogen per batch for fertilizer's A and B, respectively. Calculate the value of the test statistic.

Answer 1

`Z_(H0)`= 1.56

Explanation:

Hello!

You have two independent samples of fertilizers from distributor A and B, and need to test if the average content of nitrogen from fertilizer distributed by A is greater than the average content of nitrogen from fertilizer distributed by B.

Sample 1

X₁: Content of nitrogen of a fertilizer batch distributed by A

n₁= 4 batches

Sample mean X₁[bar]= 23pound/batch

σ₁= 4 poundes/batch

Sample 2

X₂: Content of nitrogen of a fertilizer batch distributed by B

n₂= 4 batches

Sample mean X₂[bar]= 18pound/batch

σ₂= 5 pounds/batch

Both variables have a normal distribution. Now since you have the information on the variables distribution and the values of the population standard deviations, you could use a pooled Z-test.

Your hypothesis are:

H₀: μ₁ ≤ μ₂

H₁: μ₁ > μ₂

The statistic to use is:

Z= X₁[bar] - X₂[bar] - (μ₁ - μ₂) ~N(0;1)

√(δ²₁/n₁ + δ²₂/n₂)

`Z_(H0)`= (23 - 18) - 0 = 1.56

√(16/4 + 25/4)

I hope this helps!