Question

Some researchers have speculated that the food manufacturing system in annual broadleaf seedlings might be improved with weed and soil treatments. An experiment was conducted to compare broadleaf seedling growth with soil and weeds treated with one of two herbicides. In a section of a field containing 75 seedlings, 30 were selected randomly throughout the field and assigned to receive herbicide A. The remainder received herbicide B. Soil and weeds for each seedling were treated with the appropriate herbicide, and at the end of the study period the height (in centimeters) was recorded for each seedling. The results obtained are shown.

Herbicide A: X̄₁ = 72.4 centimeters, s₁ = 9 centimeters

Herbicide B: X̄₂ = 86.1 centimeters, s₂ = 8 centimeters

Suppose we wished to determine whether there is a difference in height for the seedlings treated with the different herbicides. To answer this question, we decide to test the hypotheses H₀: μ₂ − μ₁ = 0 and H_a: μ₂ − μ₁ ≠ 0. Find the 95% confidence interval for μ₂ − μ₁ and make a conclusion based on that confidence interval.

	The 95% confidence interval is 13.7 ± 3.98. I would not reject the null hypothesis of no difference at the 0.05 level.
b	The 95% confidence interval is 13.7 ± 9.63. I would not reject the null hypothesis of no difference at the 0.05 level.
c	The 95% confidence interval is 13.7 ± 3.98. I would reject the null hypothesis of no difference at the 0.05 level.
d	The 95% confidence interval is 13.7 ± 9.63. I would reject the null hypothesis of no difference at the 0.05 level.
e	The 95% confidence interval is 13.7 ± 17.75. The p-value is less than 0.05.

Answer 1

Ans:

pooled standard deviation=SQRT(((30-1)*9^2+(45-1)*8^2)/(30+45-2))=8.4115

standard error=8.4115*SQRT((1/30)+(1/45))=1.983

df=30+45-2=73

critical t value=tinv(0.05,73)=1.993

Margin of error=1.993*1.983=3.98

point estimate=86.1-72.4=13.7

The 95% confidence interval is 13.7 ± 3.98. I would reject the null hypothesis of no difference at the 0.05 level