PRACTICE OF STATISTICS F/AP EXAM
PRACTICE OF STATISTICS F/AP EXAM
6th Edition
ISBN: 9781319113339
Author: Starnes
Publisher: MAC HIGHER
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Chapter 11.1, Problem 12E
To determine

The test for the given data at the significance level of the proposed generic model.

Expert Solution & Answer
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Answer to Problem 12E

  P -value is greater than the significance level which means that fail to reject H0 .

Explanation of Solution

Given:

    Phenotype Tall cut Tall potato Dwarf Cut Dwarfpotato
    Frequency 926288293104

Significance level: 5%=0.05

Ratio: R1:R2:W1:W2=9:3:3:1

  n=4

Concept Used:

Null hypothesis: H0

The null hypothesis is rejected when the value of P is less or equal to the significance level.

Calculation:

The null hypothesis: H0

The phenotype occur in a ratio of 9:3:3:1

  Ptc=916,Ptp=916,Pdc=316,Pdp=116

Alternative hypothesis: H1

At least one of the offspring proportion is incorrect.

Now,

Take the sum: S=9+3+3+1=16

Draw a table consisting of observed value and expected value:

The expected value is calculated as:

  E1=T×R1R=1611×916=906.19E2=T×R2R=1611×316=302.06E3=T×W1R=1611×316=302.06E4=T×W2R=1611×116=100.69

    Phenotype Tall cut Tall potato Dwarf Cut Dwarfpotato Total
    Expected value Oi9262882931041611
    Expected value Ei906.19302.06302.06100.691611

Now, use the test −statistics:

  x2=i=14(OiEi)2Eix2=(926906.19)2906.19+(288302.06)2302.06+(293302.06)2302.06+(104100.69)2100.69x2=1.468

Degree of freedom:

  df=n1=41=3

For x2=1.468 and df=3 the P -value is 0.6900 .

So, fail to reject the null hypothesis because P -value is not less or equal to, it is greater than the significance level which means that there is no difference in the distribution of generic traits.

  P>:0.6900>0.05 : fail to reject H0 .

Hence, there is no enough proof which shows that the proposed generic model of tomato plant does not obey the given standard ratio.

Conclusion:

Therefore, P>:0.6900>0.05

Chapter 11 Solutions

PRACTICE OF STATISTICS F/AP EXAM

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