Chapter 22, Problem 22.8AYK

(a)

To determine

To find:

Theexpected cell counts if the hypothesis is true of the given table.

Explanation of Solution

Given:

Landing posture	$3\text{cm}$	$5\text{cm}$	$10\text{cm}$	$20\text{cm}$
Upright	$20$	$23$	$27$	$29$
No upright	$10$	$7$	$3$	$1$
Sample size	$30$	$30$	$30$	$30$

Concept used:

Formula

  $p=\frac{x}{n}$

Significance level

  $\begin{array}{l}\alpha =1-\text{confidence}\\ \alpha =1-0.95\end{array}$

Calculation:

Parameter $p=$ sample proportion

The best point of estimate of $p$ is sample proportion

Since the sample size are large.

The condition for the two samples $z$ test are met.

Applying chi-square test of independence

  $E_i=\frac{\text{row}\text{total}\text{×column}\text{total}}{\text{grand}\text{total}}$

Draw the first table

Landing posture	$3\text{cm}$	$5\text{cm}$	$10\text{cm}$	$20\text{cm}$	Total
Upright	$24.7500$	$24.7500$	$24.7500$	$24.7500$	$99$
No upright	$5.2500$	$5.2500$	$5.2500$	$5.2500$	$21$
Sample size	$30$	$30$	$30$	$30$	$120$

(b)

To determine

To find:

Therelationship of the components in combination with the column percent of land upright dropping from $3\text{cm},5\text{cm},10\text{cm},20\text{cm}$ of the given table.

Answer to Problem 22.8AYK

  $96.667\%$

Explanation of Solution

Given:

Landing posture	$3\text{cm}$	$5\text{cm}$	$10\text{cm}$	$20\text{cm}$
Upright	$20$	$23$	$27$	$29$
No upright	$10$	$7$	$3$	$1$
Sample size	$30$	$30$	$30$	$30$

Concept used:

Formula

  $p=\frac{x}{n}$

Significance level

  $\begin{array}{l}\alpha =1-\text{confidence}\\ \alpha =1-0.95\end{array}$

Calculation:

Since the sample size are large.

The condition for the two samples $z$ test are met.

Applying chi-square test of independence

Chi-square ${\chi }^2=\frac{{\left(O_i-E_i\right)}^2}{E_i}$

Draw the second table

Landing posture	$3\text{cm}$	$5\text{cm}$	$10\text{cm}$	$20\text{cm}$	Total
Upright	$0.912$	$0.124$	$0.205$	$0.730$	$1.9697$
No upright	$4.298$	$0.583$	$0.964$	$3.440$	$9.2857$
Total	$5.2092$	$0.7071$	$1.1688$	$4.1703$	$11.2554$

Test statistic ${\chi }^2$

  $=11.255$

  $p-\text{value}=0.0104$

For aphids land upright when dropping from $20cm$

The percentage

  $\begin{array}{l}=\frac{\text{no}\text{of}\text{aphids}\text{land}\text{upright}}{n}\times 100\\ =\frac{27}{30}\times 100\\ =96.667\%\end{array}$

(c)

To determine

To explain:

The expected cell counts and chi-square relation if the hypothesis is true of the given table.

Explanation of Solution

Given:

Landing posture	$3\text{cm}$	$5\text{cm}$	$10\text{cm}$	$20\text{cm}$
Upright	$20$	$23$	$27$	$29$
No upright	$10$	$7$	$3$	$1$
Sample size	$30$	$30$	$30$	$30$

Concept used:

Formula

  $p=\frac{x}{n}$

Significance level

  $\begin{array}{l}\alpha =1-\text{confidence}\\ \alpha =1-0.95\end{array}$

Calculation:

Since the sample size are large.

The condition for the two samples $z$ test are met.

Applying chi-square test of independence

  $E_i=\frac{\text{row}\text{total}\text{×column}\text{total}}{\text{grand}\text{total}}$

Draw the first table

Landing posture	$3\text{cm}$	$5\text{cm}$	$10\text{cm}$	$20\text{cm}$	Total
Upright	$24.7500$	$24.7500$	$24.7500$	$24.7500$	$99$
No upright	$5.2500$	$5.2500$	$5.2500$	$5.2500$	$21$
Sample size	$30$	$30$	$30$	$30$	$120$

Chi-square ${\chi }^2=\frac{{\left(O_i-E_i\right)}^2}{E_i}$

Draw the second table

Landing posture	$3\text{cm}$	$5\text{cm}$	$10\text{cm}$	$20\text{cm}$	Total
Upright	$0.912$	$0.124$	$0.205$	$0.730$	$1.9697$
No upright	$4.298$	$0.583$	$0.964$	$3.440$	$9.2857$
Total	$5.2092$	$0.7071$	$1.1688$	$4.1703$	$11.2554$

Test statistic ${\chi }^2$

  $=11.255$

  $p-\text{value}=0.0104$