Question

The following table gives the yield of a hybrid variety of wheat, in quintals per
acre from 17 trial plots of land treated with four types of fertilizers
Treatment with fertilizer
A B C D
24 31 39 38
39 25 41 32
35 26 33 35
21 40 34
45 26
Estimate the number of orchards in the district.

Answer 1

Answer on Question #55137, Math Statistics and Probability

The following table gives the yield of a hybrid variety of wheat, in quintals per acre from 17 trial plots of land treated with four types of fertilizers

Treatment with fertilizer

A B C D

24 31 39 38

39 25 41 32

35 26 33 35

21 40 34

45 26

Estimate the number of orchards in the district.

Solution

We have to test $H_{o}$: $\mu_{A} = \mu_{B} = \mu_{C} = \mu_{D}$, where $\mu_{A}, \mu_{B}, \mu_{C}, \mu_{D}$ denote the mean yield per acre due to fertilizer A,B,C and respectively.

$$\mu = \frac {1}{17} \sum_ {i = 1} ^ {17} X _ {i} = 33.$$

On subtracting 333 from every observation, the given table can be written as

From the above table, we can write

$$\sum \sum X _ {i j} ^ {2} = 7 5 5; \frac {T _ {1} ^ {2}}{n _ {1}} = \frac {1}{3}; \frac {T _ {2} ^ {2}}{n _ {2}} = \frac {8 4 1}{4}; \frac {T _ {3} ^ {2}}{n _ {3}} = \frac {1 0 8 9}{5}; \frac {T _ {4} ^ {2}}{n _ {4}} = 0; \frac {T ^ {2}}{n} = \frac {9}{1 7}$$

Thus, we have

$$T S S = 7 5 5 - 9 / 1 7 = 7 5 4. 4 7$$ $$T R S S = 1 / 3 + 8 4 1 / 4 - 9 / 1 7 = 4 7 2. 8 5$$ $$E S S = T S S - T R S S = 7 5 4. 4 7 - 4 2 7. 8 5 = 3 2 6. 6 2$$

Analysis of Variance Table

From the above table, we have

$$F = 1 4 2. 6 2 / 2 5. 1 2 = 5. 6 8 > 3. 4 1$$

The critical value of $\mathrm{F}$ for 3, 13 d.f. and $\alpha = 0.05$ . Thus, $\mathrm{H}_0$ is rejected at $5\%$ level of significance.

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