7. In an agricultural experiment to determine the effects of a particular insect

Question

7. In an agricultural experiment to determine the effects of a particular insecticide, a field was planted with corn. Half the plants were sprayed with the insecticide, and half were unsprayed. Several Weeks later, independent random samples of 200 sprayed plants and 200 unsprayed plants were examined. The number of healthy plants in each sample was as follows:

                               Sprayed                 Unsprayed

Healthy                       131                        111

Unhealthy                    69                          89

If the significance level is set at 0.05, does the eviedence indicate that a higher proportion of sprayed than of unsprayed plants were healthy? Use a one tailed test (note- since the null is that the proportions are equal, use this information to construct a pooled estimate of the proportion.

Explanation / Answer

H0: The proportion of healthy sprayed plants is not significantly greater than the proportion of unsprayed healthy plants.

HA:he proportion of healthy sprayed plants is significantly greater than the proportion of unsprayed healthy plants.

s=sprayed healthhy u=unsprayed healthy

ps=.655

pu=.555

The formula for the pooled proportio is ppooled = (ps* ns + pu * nu) / (ns + nu)=(.655*200+.555*200)/(400)=.605

The standard deviation is given by {[ppooled(1-ppooled)]*(1/ns+1/np)}^1/2=.04889

The z-statistic is then (.655-.555)/.04889=2.046

Using normcdf on a TI calculator or a z-table, you find the p-value to be about .02, so you can reject the null hypothesis at the .05 significance level.

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