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PLANT X= A:1.0 B: 1.5 C: 2.0 D: 2.5 E: 3.0 F: 3.5 G: 4 H: 4.5 PLANT Y= A:3.9 B: 4.4 C: 5.8 D: 6.6 E: 7.0 F: 7.1 G: 7.3 H: 7.7 a) Plot a scatter diagram of yield y, against amount of fertilizer x. (b) calculate the equation of the least squares regression line of y on x (c) estimate the yield of a plant treated weekly with 3.2grams of fertilizer. (d) explain why it may not be appropriate t use your equation to predict the yield of a tomato plant treated weekly with 20grams of fertilizer. QUESTION 3 Students were given different drug treatment before revising for their examinations. some were given a memory drug, some a placebo and others no treatment. the exam scores (%) are shown for the three groups: MEMORY DRUG: 70,77,83,90,97. Mean: 83.40, variance: 112.40 PLACEBO: 37,43,50, 57, 63. mean:50.00, variance: 109.00 NO TREATMENT: 2, 10, 17, 23, 30 . Mean: 16.60, Variance: 112.30 GRAND MEAN: 50.00 GRAND VARIANCE: 892.14 (A) Carry out an analysis of variance to determine wether the treatments have different effects on the performance of the student (use alpha = 0.05)

Question

PLANT X= A:1.0 B: 1.5 C: 2.0 D: 2.5 E: 3.0 F: 3.5 G: 4 H: 4.5 PLANT Y= A:3.9 B: 4.4 C: 5.8 D: 6.6 E: 7.0 F: 7.1 G: 7.3 H: 7.7 a) Plot a scatter diagram of yield y, against amount of fertilizer x. (b) calculate the equation of the least squares regression line of y on x (c) estimate the yield of a plant treated weekly with 3.2grams of fertilizer. (d) explain why it may not be appropriate t use your equation to predict the yield of a tomato plant treated weekly with 20grams of fertilizer. QUESTION 3 Students were given different drug treatment before revising for their examinations. some were given a memory drug, some a placebo and others no treatment. the exam scores (%) are shown for the three groups: MEMORY DRUG: 70,77,83,90,97. Mean: 83.40, variance: 112.40 PLACEBO: 37,43,50, 57, 63. mean:50.00, variance: 109.00 NO TREATMENT: 2, 10, 17, 23, 30 . Mean: 16.60, Variance: 112.30 GRAND MEAN: 50.00 GRAND VARIANCE: 892.14 (A) Carry out an analysis of variance to determine wether the treatments have different effects on the performance of the student (use alpha = 0.05)

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Solution

This is a complex question that involves multiple parts. Let's break it down:

a) To plot a scatter diagram of yield y, against the amount of fertilizer x, you would need to use a graphing tool. On the x-axis, you would plot the amount of fertilizer (from A to H for both Plant X and Plant Y). On the y-axis, you would plot the yield. Each point on the graph represents a pair of (fertilizer, yield) for each plant.

b) To calculate the equation of the least squares regression line of y on x, you would need to use the formula for the least squares method. This involves calculating the slope (b1) and the y-intercept (b0) using the given x and y values. The formula for b1 is [NΣXY - (ΣX)(ΣY)] / [NΣX^2 - (ΣX)^2] and for b0 is [ΣY - b1(ΣX)] / N.

c) To estimate the yield of a plant treated weekly with 3.2 grams of fertilizer, you would need to substitute x = 3.2 into the equation of the regression line obtained in part b.

d) It may not be appropriate to use your equation to predict the yield of a tomato plant treated weekly with 20 grams of fertilizer because the data provided only goes up to 4.5 grams of fertilizer. Predicting a yield for 20 grams of fertilizer would be an extrapolation, which can be unreliable as it goes beyond the range of the data.

For the second question, an analysis of variance (ANOVA) is used to determine whether there are any statistically significant differences between the means of three or more independent groups. In this case, the groups are the students who were given a memory drug, a placebo, and no treatment. The steps to perform an ANOVA include calculating the sum of squares between (SSB), sum of squares within (SSW), total sum of squares (SST), degrees of freedom between (dfB), degrees of freedom within (dfW), mean square between (MSB), mean square within (MSW), and finally the F statistic. The F statistic is then compared to the critical value from the F-distribution table to determine if the treatments have different effects on the performance of the students.

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