Question

Question 3. In a study measuring the height of birch trees, two types of fertilizer were used. Prior to the study 14 tree's heights were measured. 8 trees were then selected at random to receive a liquid fertilizer during the first summer of the experiment. At the end of the first year their growth was measured. While this was occurring another 6 trees were receiving a granular fertilizer. At the end first year their height was measured. During year 2 the trees that had received the granular fertilizer now received the liquid fertilizer and vice versa. At the end of the second year the heights of the plants was recorded. Trees Measurement (Feet) Granular First Year 0 Year 1 Year 1 minus 0 Year 2 Year 2 minus 1 4.6 4.8 0.2 5.3 0.5 2 4.8 4.9 0.1 5.4 0.5 3 3.9 4.3 0.4 4.7 0.4 4 4.2 4.5 0.3 4.8 0.3 5 4.8 5.1 0.3 5.7 0.6 6 4.7 4.8 0.1 5.2 0.4 1 Nwl Liquid First 4.2 4.7 0.5 4.9 0.2 4.5 5.1 0.6 5.3 0.2 3 3.8 3.9 0.1 4.0 0.1 4 5.0 5.5 0.5 5.7 0.2 5 4.9 5.6 0.7 5.7 0.1 6 4.7 5.2 0.5 5.5 0.3 7 4.6 4.9 0.3 5.2 0.3 8 4.5 5.1 0.6 5.3 0.2 Test whether the liquid fertilizer results in greater growth than granular fertilizer. Be sure to include the 4 steps of an hypothesis test.

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Answer 1

Sol:

To test if the liquid fertilizer results in greater growth than in granular fertilizer, let uz be the average growth rate of liquid fertilizer and yo be the average growth rate of granular fertilizer. The hypothesis is formulated below: Họ: A. = 4G H:Hy > HG Compute the average growth rates for both the fertilizers as shown in the table below: Liquid (1-0) (2-1) 0.5 0.6 0.1 0.5 0.7 0.5 0.3 0.6 Average 0.2 0.35 0.2 0.4 0.1 0.1 0.2 0.35 0.1 0.4 0.3 0.4 0.3 0.3 0.2 0.4

Granular (1-0) (2-1) 0.2 Average 0.5 0.35 0.5 0.3 0.1 0.4 0.4 0.3 0.4 0.3 0.3 0.3 0.6 0.45 0.25 0.1 0.4 Now, compare the two-sample means, using a two-sample t-test for independent samples: Average Liquid Average Granular 0.35 0.35 0.40 0.30 0.10 0.40 0.35 0.30 0.40 0.45 0.40 0.25 0.30 0.40 Compute the sample mean and standard deviation using the formula below:

0.35 +0.40+...+0.30 8 = 0.337 0.35 +0.30+...+0.25 = 6 = 0.3417 (x, – F;) i=1 SL = n-1 = 0.103 8 (x; -c) i=1 SG - n-1 = 0.0736 Compute the test statistic using formula below: t= iskt X; - XG sã no no 0.337 -0.3417 0.1032 0.0736 + 8 6 = -0.099 Degrees of freedom n + n - 2 = 8+6-2 =12 p-value = 0.539 (Using MS-Excel function 1-T.DIST(-0.099,12, TRUE))

As the p-value is greater than 0.05, so the null hypothesis will not be rejected at 5% level of significance. Hence, it is concluded that there is not sufficient evidence to support the claim that liquid fertilizer results in greater growth than in granular fertilizer.

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