Question

Charlotte buys $20$ strawberry plants for her garden. She feeds half each day with water and the other half with a special strawberry plant food. After one month she records the amount of strawberries on each plant.

Number of strawberries on each plant fed with water
$15$	$13$	$12$	$15$	$17$	$16$	$12$	$10$	$11$	$15$

Number of strawberries on each plant fed with special strawberry plant food
$14$	$15$	$17$	$17$	$19$	$19$	$12$	$14$	$15$	$15$

Analyze the data to find whether there is an effect of using the special strawberry plant food.

Accepted Answer

Consider the table of:

Number of strawberries on each plant fed with water,

Number of strawberries on each plant fed with water $(x_{i})$ $(x_{i} - \bar{x})$ ${(x_{i} - \bar{x})}^{2}$

$15$ $1.4$ $1.96$

$13$ $- 0.6$ $0.36$

$14$ $15$ $2.56$

$15$ $1.4$ $1.96$

$17$ $3.4$ $11.56$

$16$ $2.4$ $5.76$

$12$ $- 1.6$ $2.56$

$10$ $- 3.6$ $12.96$

$11$ $2.6$ $6.76$

$15$ $1.4$ $1.96$

$\sum x_{i} = 136$ $\sum {(x_{i} - \bar{x})}^{2} = 48.4$

Number of strawberry plants $n = 10$ .

Mean $\bar{x} = \frac{\sum x_{i}}{n} = \frac{136}{10} = 13.6$

Standard deviation $σ_{x} = \sqrt{\frac{\sum {(x_{i} - \bar{x})}^{2}}{n}} = \sqrt{\frac{48.4}{10}} = 2.2$

Consider the table of:

Number of strawberries on each plant fed with special strawberry plant food,

Number of strawberries on each plant fed with special strawberry plant food $(y_{i})$ $(y_{i} - \bar{y})$ ${(y_{i} - \bar{y})}^{2}$

$14$ $- 1.7$ $2.89$

$15$ $- 0.7$ $0.49$

$17$ $1.3$ $1.69$

$17$ $1.3$ $1.69$

$19$ $3.3$ $10.89$

$19$ $3.3$ $10.89$

$12$ $- 3.7$ $13.69$

$14$ $- 1.7$ $2.89$

$15$ $- 0.7$ $0.49$

$15$ $- 0.7$ $0.49$

$\sum y_{i} = 157$ $\sum {(y_{i} - \bar{y})}^{2} = 46.1$

Number of strawberry plants $n = 10$ .

Mean $\bar{y} = \frac{\sum y_{i}}{n} = \frac{157}{10} = 15.7$

Standard deviation $σ_{y} = \sqrt{\frac{\sum {(y_{i} - \bar{y})}^{2}}{n}} = \sqrt{\frac{46.1}{10}} = 2.15$

Here $\bar{y} > \bar{x}$ and $σ_{y} < σ_{x}$

So, plant fed with special strawberry plant food is better than the plant fed with water because mean is higher and standard deviation is lower.

Day	Food $(g)$
$1$	$55$
$2$	$64$
$3$	$45$
$4$	$54$
$5$	$60$
$6$	$50$
$(y_{i} - \bar{y})$	$59$
$8$	$61$
$9$	$49$

Day	Food $(g)$
$1$	$55$
$2$	$64$
$3$	$45$
$4$	$54$
$5$	$60$
$6$	$50$
$7$	$59$
$8$	$61$
$9$	$49$

Day	Food $(g)$
$1$	$55$
$2$	$0.49$
$3$	$45$
$4$	$54$
$5$	$60$
$6$	$50$
$7$	$59$
$8$	$61$
$9$	$49$

$6.3$	$9.6$	$12.2$	$12.3$	$10.3$
$12.1$	$10.3$	$8.4$	$9.2$	$4.3$

$6.3$	$9.6$	$12.2$	$12.3$	$10.3$
$12.1$	$10.3$	$8.4$	$9.2$	$4.3$

Rose Harrison, Clara Huizink, Aidan Sproat Clements and, Marlene Torres Skoumal Solutions for Exercise 18: Mixed practice

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