EBK PRACTICE OF STATISTICS IN THE LIFE
EBK PRACTICE OF STATISTICS IN THE LIFE
4th Edition
ISBN: 9781319067496
Author: BALDI
Publisher: VST
Question
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Chapter 22, Problem 22.8AYK

(a)

To determine

To find:

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

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

    Landing posture3cm5cm10cm20cm
    Upright 20232729
    No upright10731
    Sample size30303030

Concept used:

Formula

  p=xn

Significance level

  α=1confidenceα=10.95

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

  Ei=rowtotal×columntotalgrandtotal

Draw the first table

    Landing posture3cm5cm10cm20cmTotal
    Upright 24.750024.750024.750024.750099
    No upright5.25005.25005.25005.250021
    Sample size30303030120

(b)

To determine

To find:

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

(b)

Expert Solution
Check Mark

Answer to Problem 22.8AYK

  96.667%

Explanation of Solution

Given:

    Landing posture3cm5cm10cm20cm
    Upright 20232729
    No upright10731
    Sample size30303030

Concept used:

Formula

  p=xn

Significance level

  α=1confidenceα=10.95

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 χ2=(OiEi)2Ei

Draw the second table

    Landing posture3cm5cm10cm20cmTotal
    Upright 0.9120.1240.2050.7301.9697
    No upright4.2980.5830.9643.4409.2857
    Total 5.20920.70711.16884.170311.2554

Test statistic χ2

  =11.255

  pvalue=0.0104

For aphids land upright when dropping from 20cm

The percentage

  =noofaphidslanduprightn×100=2730×100=96.667%

(c)

To determine

To explain:

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

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

    Landing posture3cm5cm10cm20cm
    Upright 20232729
    No upright10731
    Sample size30303030

Concept used:

Formula

  p=xn

Significance level

  α=1confidenceα=10.95

Calculation:

Since the sample size are large.

The condition for the two samples z test are met.

Applying chi-square test of independence

  Ei=rowtotal×columntotalgrandtotal

Draw the first table

    Landing posture3cm5cm10cm20cmTotal
    Upright 24.750024.750024.750024.750099
    No upright5.25005.25005.25005.250021
    Sample size30303030120

Chi-square χ2=(OiEi)2Ei

Draw the second table

    Landing posture3cm5cm10cm20cmTotal
    Upright 0.9120.1240.2050.7301.9697
    No upright4.2980.5830.9643.4409.2857
    Total 5.20920.70711.16884.170311.2554

Test statistic χ2

  =11.255

  pvalue=0.0104

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