Question

A test is conducted on 25 lemon trees of the same species to compare two fertilizers, A and B. A sample of 13 lemon trees randomly selected from the group are given fertilizer A and the remaining trees are given fertilizer B. From observations made over a three-week period, the average fruits produced by each lemon tree is recorded below:

Lemon Fruit Yield (for each tree)
fertilizer A (x1) 44 44 56 46 47 38 58 53 49 35 46 30 41
fertilizer B (x2) 35 47 55 29 40 39 32 41 42 57 51 39

Assume these two samples come from independent normally distributed populations.

6) Find the mean number of lemons for fertilizer A rounded to the tenths place.

7) Find the mean number of lemons for fertilizer A rounded to the tenths place.

Now perform the two-sample t-test to investigate the evidence of a difference in true mean lemon yields for the two fertilizers.

8) Using a pooled variance, find the p-value.

9) Which of the following hypotheses are most appropriate in determining whether there is a difference in the true mean lemon yields for the two fertilizers? (Choose 2)

• Ho: μ1 = μ2

• Ho: μ1 ≠ μ2

• Ha: μ1 = μ2

• Ha: μ1 ≠ μ2

• Ho: μ1 > μ2

• Ho: μ1< μ2

• Ha: μ1 > μ2

• Ha: μ1 < μ2

Answer 1

ANSWER::

6)

For Fertilizer A :

∑x = 587

∑x² = 27273

n1 = 13

Mean , x̅1 = Ʃx/n = 587/13 = 45.1538 = 45.15

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(27273-(587)²/13)/(13-1)] = 7.9984

7)

For Fertilizer B :

∑x = 507

∑x² = 22261

n2 = 12

Mean , x̅2 = Ʃx/n = 507/12 = 42.25

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(22261-(507)²/12)/(12-1)] = 8.7399

8)

Pooled variance :

S²p = ((n1-1)*s1² + (n2-1)*s2² )/(n1+n2-2) = ((13-1)*7.9984² + (12-1)*8.7399²) / (13+12-2) = 69.9105

9)

Null and Alternative hypothesis:

Ho : µ1 = µ2

H1 : µ1 ≠ µ2

Test statistic:

t = (x̅1 - x̅2) / √(s²p(1/n1 + 1/n2 ) = (45.1538 - 42.25) / √(69.9105*(1/13 + 1/12)) = 0.8676

df = n1+n2-2 = 23

p-value = T.DIST.2T(ABS(0.8676), 23) = 0.3946

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