Question

Some researchers have conjectured that stem-pitting disease in peach-tree seedlings might be controlled with weed and soil treatment. An experiment is conducted to compare peach-tree seedling growth when the soil and weeds are treated with one of two herbicides. In a field containing 20 seedlings, 10 are randomly selected throughout the field and assigned to receive herbicide A. The remainder of the seedlings is assigned to receive herbicide B. Soil and weeds for each seedling are treated with the appropriate herbicide, and at the end of the study period, the height in centimeters is recorded for each seedling. The following results are obtained: Herbicide A = 94.5 cm s1 = 10 cm Herbicide B = 109.1 cm s2 = 9 cm
What is a 90% confidence interval (use the conservative value for the degrees of freedom) for μ2 – μ1? a. 14.6 ± 7.00 b. 14.6 ± 7.38 c. 14.6 ± 7.80 d. 14.6 ± 9.62

Suppose we wish to determine if there tends to be 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, Ha: μ2 – μ1 ≠ 0. What is the value of the two-sample t statistic? a. 14.6 b. 3.43 c. 2.54 d. 7.80

What can we say about the value of the P-value? a. P-value < 0.01 b. 0.01 < P-value < 0.05 c. 0.05 < P-value < 0.10 d. P-value > 0.10

If you use the 0.05 level of significance, what conclusion would you reach? a. The evidence is not sufficiently strong to reject the null hypothesis. b. There is sufficient evidence to reject the null hypothesis. c. Cannot be determined by the information given.

Answer 1

Answer:

Part A

b. 14.6 ± 7.38

Part B

b. 3.43

Part C

a. P-value < 0.01

Part D

b. There is sufficient evidence to reject the null hypothesis

Step-by-step explanation:

Part A

The given data are;

The number of seedlings in the field = 20

The number of seedlings selected to receive herbicide A = 10

The number of seedlings selected to receive herbicide B = 10

The height in centimeters of seedlings treated with Herbicide A, $\overline x _1$ = 94.5 cm

The standard deviation, s₁ = 10 cm

The height in centimeters of seedlings treated with Herbicide B, $\overline x _2$ = 109.1 cm

The standard deviation, s₂ = 9 cm

The 90% confidence interval for μ₂ - μ₁, is given as follows;

$\left (\bar{x}_{2}- \bar{x}_{1} \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}$

The critical-t at 95% and n₁ + n₂ - 2 degrees of freedom is given as follows;

The degrees of freedom, df = n₁ + n₂ - 2 = 10 + 10 - 2 = 18

α = 100% - 90% = 10%

∴ For two tailed test, we have, α/2 = 10%/2 = 5% = 0.05

$t_{(0.025, \, 18)}$ = 1.734

$C.I. = \left (109.1- 94.5 \right )\pm 1.734 \times \sqrt{\dfrac{10^{2}}{10}+\dfrac{9^{2}}{10}}$

C.I. ≈ 14.6 ± 7.37714603353

The 90% C.I. ≈ 14.6 ± 7.38

b. 14.6 ± 7.38

Part B

With the hypotheses are given as follows;

H₀; μ₂ - μ₁ = 0

Hₐ; μ₂ - μ₁ ≠ 0

The two sample t-statistic is given as follows;

$t=\dfrac{(\bar{x}_{2}-\bar{x}_{1})}{\sqrt{\dfrac{s_{1}^{2} }{n_{1}}+\dfrac{s _{2}^{2}}{n_{2}}}}$

$t-statistic=\dfrac{(109.1-94.5)}{\sqrt{\dfrac{10^{2} }{10}+\dfrac{9^{2}}{10}}} \approx 3.43173361147$

The two sample t-statistic ≈ 3.43

b. 3.43

Part C

From the t-table, the p-value, we have, the p-value < 0.01

a. P-value < 0.01

Part D

Given that a significance level of 0.05 level is used and the p-value of 0.01 is less than the significance level, there is enough statistical evidence to reject the null hypothesis

b. There is sufficient evidence to reject the null hypothesis.