Question

A typical lawn averages 21 blades of grass per square inch. A garden store sells an expensive fertilizer designed to transform an average lawn into a thick carpet within three weeks. To test the claim, a random sample of 32 average lawn plots were selected and treated with the fertilizer. Three weeks later the density of each plot was measured by counting the blades of grass per square inch. n = 32 and X = 22.4 Based on data from previous studies assume a is known to be 2.7. Do we have significant evidence the fertilizer increases the grass density?

32a

Write the null and alternative hypothesis.

32b

a = .01. Draw a picture of the rejection region on the appropriate curve.

32c

Compute the test statistic.

32d

What is your conclusion?

32e

Draw a picture and find the p-value.

Hint: remember this rule: H0 is rejected iff p-value < a

Answer 1

a) H₀: = 21

    H₁: > 21

b) At alpha = 0.01, the critical value is z_0.99 = 2.33

c) The test statistic z = ()/()

                                  = (22.4 - 21)/(2.7/)

                                  = 2.93

d) Since the test statistic value is greater than the critical value(2.93 > 2.33), so we should reject the null hypothesis.

So at 0.01 significance level there is sufficient evidence to conclude that the fertilizer increases the grass density.

e) P-value = P(Z > 2.93)

                 = 1 - P(Z < 2.93)

                 = 1 - 0.9983

                 = 0.0017

Since the P-value is less than the significance level(0.0017 < 0.01), so we should reject H₀.