Question

Two equal groups of seedlings, and equal in height, were selected for an experiment. One group of seedlings was fed Fertilizer A, and the other group with Fertilizer B. The mean heights of the two groups of seedlings, their standard deviations, and sample sizes are listed below. Assume simple random sampling, independence, and normally distributed data.Alpha = 0.025.Sample Mean Height (inches, Std Deviation (inches), Sample SizeFertilizer A 12.92, 0.25, 15 Fertilizer B 12.63, 0.20, 13What is the null hypothesis?a. Using the data in the previous question, what is the critical t value using the simplified textbook method and Tables when testing the claim that Fertilizer A is significantly greater than Fertilizer B? Record the answer to three decimal places (x.xxx).b.Using the data from the previous question, what is the t-statistic rounded off to one decimal place ? (x.x)c. Using the data from above, Fertilizer A is significantly greater than Fertilizer B using an alpha of 0.025?

Answer 1

Answer:

Null Hypothesis: H_0: \mu_A =\mu _B or \mu_A -\mu _B=0

Alternate Hypothesis: H_1: \mu_A >\mu _B or \mu_A -\mu _B>0

Here to test Fertilizer A height is greater than Fertilizer B

Two Sample T Test:

t=\frac{X_1-X_2}{\sqrt{S_p^2(1/n_1+1/n_2)}}

Where S_p^2=\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}

S_p^2=\frac{(14)0.25^2+(12)0.2^2}{15+13-2}= 0.0521154

t=\frac{12.92-12.63}{\sqrt{0.0521154(1/15+1/13)}}= 3.3524

P value for Test Statistic of P(3.3524,26) = 0.0012

df = n1+n2-2 = 26

Critical value of P : t_{0.025,26}=2.05553

We can conclude that Test statistic is significant. Sufficient evidence to prove that we can Reject Null hypothesis and can say Fertilizer A is greater than Fertilizer B.