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 µ1and µ2 represent the average amount of nitrogen per batch for fertilizer’s A and B, respectively. Which of the following is the appropriate conclusion at the 5% significance level? The test statistic calculated in Excel with these data is 1.5617.

Answer 1

Answer:

$z=\frac{(23-18)-0}{\sqrt{\frac{4^2}{4}+\frac{5^2}{4}}}}=1.5617$  

$p_v =P(Z>1.5617)=0.059$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the difference between the true mean of group A and B is not significantly higher than 0 at 5% of significance.

Step-by-step explanation:

Data given and notation

$\bar X_{A}=23$ represent the mean for the sample A

$\bar X_{B}=18$ represent the mean for the sample B

$\sigma_{A}=4$ represent the population standard deviation for the sample A

$\sigma_{B}=5$ represent the population standard deviation for the sample B

$n_{A}=4$ sample size selected A

$n_{B}=4$ sample size selected B

$\alpha=0.05$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean for A is higher than the mean for B, the system of hypothesis would be:

Null hypothesis:$\mu_{A}-\mu_{B}\leq 0$

Alternative hypothesis:$\mu_{A}-\mu_{B}>0$

We know the population deviations, so for this case is better apply a z test to compare means, and the statistic is given by:

$z=\frac{(\bar X_{A}-\bar X_{B})-0}{\sqrt{\frac{\sigma^2_{A}}{n_{A}}+\frac{\sigma^2_{B}}{n_{B}}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{(23-18)-0}{\sqrt{\frac{4^2}{4}+\frac{5^2}{4}}}}=1.5617$  

P-value

Since is a one right tailed test the p value would be:

$p_v =P(Z>1.5617)=0.059$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the difference between the true mean of group A and B is not significantly higher than 0 at 5% of significance.