Question

A biologist was interested in determining whether sunflower seedlings treated with an extract from Vinca minor roots resulted in a lower average height of sunflower seedlings than the standard height of 15.7 cm (σ = 3.5 cm). The biologist treated a random sample of n = 31 seedlings with the extract and subsequently obtained the sample mean: 13.8 cm. Use a 0.01 level of significance to test the hypothesis that μ = 15.7 (H0) against the alternative that μ < 15.7 (H1).

Answer 1

Solution:

Given:

Average standard height = cm

Standard Deviation =

Sample size = n = 31

Sample mean =

level of significance = 0.01

Null Hypothesis:

H0: μ = 15.7 (H0) against

Alternative Hypothesis:

H1: μ < 15.7 ( Left tailed test)

Find test statistic value:

Find z critical value:

level of significance = 0.01 and this is left tailed test, thus look in z table for area = 0.0100 or its closest area and find z value.

Area 0.0099 is closest to 0.0100 and it corresponds to -2.3 and 0.03

thus z critical value = -2.33

Decision Rule:
Reject null hypothesis ,if z  test statistic value < z critical value = -2.33 , otherwise we fail to reject H0.

Since z  test statistic value = z = -3.02 < z critical value = -2.33 ,

Conclusion:

At 0.01 level of significance, we have sufficient evidence to conclude that sunflower seedlings treated with an extract from Vinca minor roots resulted in a lower average height of sunflower seedlings than the standard height of 15.7 cm.