Chapter 22, Problem 22.4AYK

(a)

To determine

To find:

Thepercent of aphids land upright dropping from $3\text{cm},5\text{cm},10\text{cm},20\text{cm}$ of the given table.

Answer to Problem 22.4AYK

  $66.667\%$

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:

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.

The population distribution is normal.

Sample size $n=30$

For aphids land upright when dropping from $3cm$

The percentage

  $\begin{array}{l}=\frac{\text{no}\text{of}\text{aphids}\text{land}\text{upright}}{n}\times 100\\ =\frac{20}{30}\times 100\\ =66.667\%\end{array}$

For aphids land upright when dropping from $5cm$

The percentage

  $\begin{array}{l}=\frac{\text{no}\text{of}\text{aphids}\text{land}\text{upright}}{n}\times 100\\ =\frac{23}{30}\times 100\\ =76.667\%\end{array}$

For aphids land upright when dropping from $10cm$

The percentage

  $\begin{array}{l}=\frac{\text{no}\text{of}\text{aphids}\text{land}\text{upright}}{n}\times 100\\ =\frac{29}{30}\times 100\\ =90\%\end{array}$

For aphids land upright when dropping from $20cm$

The percentage

  $\begin{array}{l}=\frac{\text{no}\text{of}\text{aphids}\text{land}\text{upright}}{n}\times 100\\ =\frac{29}{30}\times 100\\ =96.667\%\end{array}$

(b)

To determine

To explain:

The $95\%$ confident are not capture the respective population proportion 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.

The population distribution is normal.

Sample size $n=30$

For aphids land upright when dropping from $3cm$

The percentage

  $\begin{array}{l}=\frac{\text{no}\text{of}\text{aphids}\text{land}\text{upright}}{n}\times 100\\ =\frac{20}{30}\times 100\\ =66.667\%\end{array}$

For aphids land upright when dropping from $5cm$

The percentage

  $\begin{array}{l}=\frac{\text{no}\text{of}\text{aphids}\text{land}\text{upright}}{n}\times 100\\ =\frac{23}{30}\times 100\\ =76.667\%\end{array}$

For aphids land upright when dropping from $10cm$

The percentage

  $\begin{array}{l}=\frac{\text{no}\text{of}\text{aphids}\text{land}\text{upright}}{n}\times 100\\ =\frac{29}{30}\times 100\\ =90\%\end{array}$

For aphids land upright when dropping from $20cm$

The percentage

  $\begin{array}{l}=\frac{\text{no}\text{of}\text{aphids}\text{land}\text{upright}}{n}\times 100\\ =\frac{29}{30}\times 100\\ =96.667\%\end{array}$

For aphids land upright dropping from $20cm$ is the percent of $96.667\%$ .

This percent is large of the $95\%$ confident.

So,the $95\%$ confident are not capture the respective population proportion of all four intervals.

(c)

To determine

To explain:

The $95\%$ confident are not capture the respective population proportion 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.

The population distribution is normal.

Sample size $n=30$

For aphids land upright when dropping from $3cm$

The percentage

  $\begin{array}{l}=\frac{\text{no}\text{of}\text{aphids}\text{land}\text{upright}}{n}\times 100\\ =\frac{20}{30}\times 100\\ =66.667\%\end{array}$

For aphids land upright when dropping from $5cm$

The percentage

  $\begin{array}{l}=\frac{\text{no}\text{of}\text{aphids}\text{land}\text{upright}}{n}\times 100\\ =\frac{23}{30}\times 100\\ =76.667\%\end{array}$

For aphids land upright when dropping from $10cm$

The percentage

  $\begin{array}{l}=\frac{\text{no}\text{of}\text{aphids}\text{land}\text{upright}}{n}\times 100\\ =\frac{29}{30}\times 100\\ =90\%\end{array}$

For aphids land upright when dropping from $20cm$

The percentage

  $\begin{array}{l}=\frac{\text{no}\text{of}\text{aphids}\text{land}\text{upright}}{n}\times 100\\ =\frac{29}{30}\times 100\\ =96.667\%\end{array}$

Null hypothesis $H_0$ :-No relationship between aphids land upright and landing posture.

Alternative hypothesis $H_1$ :-A significant relationship between aphids land upright and landing posture.

Significance level

  $\begin{array}{l}\alpha =1-\text{confidence}\\ \alpha =1-0.95\\ \alpha =0.05\end{array}$

Since $p-$ Value is greater than $0.05$ significance level.

So the hypothesis is fail to reject null hypothesis $H_0$