PRACTICE OF STATS - 1 TERM ACCESS CODE
PRACTICE OF STATS - 1 TERM ACCESS CODE
4th Edition
ISBN: 9781319403348
Author: BALDI
Publisher: Macmillan Higher Education
Question
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Chapter 21, Problem 21.22E

(a)

To determine

To explain:

Theconditions for a chi-square goodness of fit test of the given law.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

Tall plants =787

Dwarf plants =277

Concept used:

Formula

Degree of freedom

  =n1

Significance level

  α=1confidenceα=10.95

Calculation:

The sample is random.

The variable under steady is tall or dawn f .

The expected value of number of observations in each level of the variable is at least 5 .

All the conditions for chi-square goodness of fit test satisfied.

Since the sample size are large.

The condition for the two samples z test are met.

(b)

To determine

To explain:

Thenull and alternative hypothesis fir this test of the given law.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

Tall plants =787

Dwarf plants =277

Concept used:

Formula

Degree of freedom

  =n1

Significance level

  α=1confidenceα=10.95

Calculation:

Since the sample size are large.

The condition for the two samples z test are met.

Null hypothesis H0 :-No relationship between expected ratio and observed ratio.

Alternative hypothesis H1 :-A significant relationship between expected ratio and observed ratio Significance level

  α=1confidenceα=10.95α=0.05

Since p Value is greater than 0.05 significance level.

So the hypothesis is fail to reject null hypothesis H0

(c)

To determine

To find:

The expected counts under the null hypothesis test of the given law.

(c)

Expert Solution
Check Mark

Answer to Problem 21.22E

  798 , 266

Explanation of Solution

Given:

Tall plants =787

Dwarf plants =277

Concept used:

Formula

Degree of freedom

  =n1

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

E (Tall)

  =34×1064=798

E (short)

  =14×1064=266

(d)

To determine

To find:

The chi-square statistic and the p-valueof the given law.

(d)

Expert Solution
Check Mark

Answer to Problem 21.22E

The P-value =0.4361

Explanation of Solution

Given:

Tall plants =787

Dwarf plants =277

Concept used:

Formula

Degree of freedom

  =n1

Significance level

  α=1confidenceα=10.95

Calculation:

  H0:π1=π2=π3

  H1:notalloftheπjareequal

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

  χ2=(OiEi)2Eiχ2=(787798)2798+(277266)2266χ2=0.6065

Chi-square =0.6065

Degree of freedom

  =n1=(21)=1

P-value =0.4361

This result is not significant.

So, the data is not support the Mendel’s first law.

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