# Does the New Spray Reduce Loss Due to Insect Damage?

## A fruit grower wants to test a new spray that a manufacturer claims will reduce the loss due to insect damage.

The grower sprayed 200 trees with the new spray and 200 other trees with the standard spray. The mean yield per tree for the new spray was 109 and for the standard spray was 103. The variance for the new spray was 372.

Do the data provide sufficient evidence to conclude that the mean yield per tree treated with the new spray is more than that for trees treated with the standard spray? Use a = 0.01. Use the p-value approach.

## Question:

Do the data suggest that the new spray is more effective in reducing loss due to insect damage compared to the standard spray?

## Answer:

To determine if the new spray is more effective, a statistical test must be performed. A Z-test will allow us to evaluate if the differences in yield are statistically significant. If the p-value from the test is less than our level of significance (0.01), we could conclude that the new spray is indeed more effective.

**Explanation:** The subject matter here is statistical hypothesis testing. The fruit grower used both the new and the standard spray on different sets of trees and recorded the yield per tree for each. Looking at the problem, it seems like we are setting up for a test of two population means with known variances, so we will use a Z-test. With all hypothesis tests, we start with a null and alternative hypothesis. The null hypothesis will be that the new spray does not improve the yield per tree over the standard spray, i.e. mean(new spray) = mean(standard spray). The alternative hypothesis is what the manufacturer claims, that the new spray does improve the yield per tree, i.e. mean(new spray) > mean(standard spray).

Our Z-score will then tell us whether or not to reject the null hypothesis. If the P-value is less than our alpha level (0.01), we reject the null hypothesis and conclude that there is statistical evidence to support the manufacturer's claim. On the other hand, if the P-value is greater than alpha, we do not reject the null hypothesis, suggesting that the new spray does not significantly improve the yield per tree over the standard spray when it comes to reducing insect damage.

To calculate the Z-score you would use the following formula: Z = (x – μ) / (σ/√n) and for the p-value you would look at the appropriate value in the Z-distribution table. Trees treated with the new spray should show a statistically different outcome than those sprayed with the standard. But we won’t know for sure until the statistical test is done.