Practice of Statistics in the Life Sciences
Practice of Statistics in the Life Sciences
4th Edition
ISBN: 9781319013370
Author: Brigitte Baldi, David S. Moore
Publisher: W. H. Freeman
Question
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Chapter 22, Problem 22.12AYK

(a)

To determine

To explain:

The conditions are very similar of the given data.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

  x1=177,x2=86n1=434,n2=222

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.

To met chi square test, the population distribution is normal.

  p=xn

(b)

To determine

To find:

The value of the statistic p of the given sample.

(b)

Expert Solution
Check Mark

Answer to Problem 22.12AYK

  p Value =0.6136

Explanation of Solution

Given:

  x1=177,x2=86n1=434,n2=222

Concept used:

Formula

Standard of error

  SE=p(1p)(1n1+1n2)

Significance level

  α=1confidenceα=10.95

Calculation:

Parameter p= true proportion of individuals in population

The best point of estimate of p is sample proportion

  X Is a random variable

  H0:p1=p2H1:p1>p2

  p^1=177434p^1=0.4078p^2=86222p^2=0.3874

The value of p is

  p^N(p,pqn)

  p Comes from null hypothesis and alternative hypothesis

  z=p^ppqn

Pooled proportion

  p=177+86434+222p=0.4009146

Standard of error

  SE=p(1p)(1n1+1n2)SE=0.4009146(10.4009146)(1434+1222)SE=0.0404

  z=p^1p^2SEz=0.40780.38740.0404z=0.5050

For two-tail test

  p Value

  =2P(z>z0)=2P(z>0.5050)=0.6136

(c)

To determine

To find:

The value of the chi-square χ2 of the given sample.

(c)

Expert Solution
Check Mark

Answer to Problem 22.12AYK

  χ2=0.5050

Explanation of Solution

Given:

  x1=177,x2=86n1=434,n2=222

Concept used:

Formula

Standard of error

  SE=p(1p)(1n1+1n2)

Significance level

  α=1confidenceα=10.95

Calculation:

Parameter p= true proportion of individuals in population

The best point of estimate of p is sample proportion

  X Is a random variable

  H0:p1=p2H1:p1>p2

  p^1=177434p^1=0.4078p^2=86222p^2=0.3874

The value of p is

  p^N(p,pqn)

  p Comes from null hypothesis and alternative hypothesis

  z=p^ppqn

Pooled proportion

  p=177+86434+222p=0.4009146

Standard of error

  SE=p(1p)(1n1+1n2)SE=0.4009146(10.4009146)(1434+1222)SE=0.0404

  χ2=p^1p^2SEχ2=0.40780.38740.0404χ2=0.5050

  p Value =0.6136

Since p Value is greater than 0.05 significance level.

So the hypothesis is fail to reject null hypothesis H0

(d)

To determine

To explain:

The effectiveness of supplementing t-PA with endovascular therapy in patients with an acute ischemic strokeof the given data.

(d)

Expert Solution
Check Mark

Explanation of Solution

Given:

  x1=177,x2=86n1=434,n2=222

Concept used:

Formula

Standard of error

  SE=p(1p)(1n1+1n2)

Significance level

  α=1confidenceα=10.95

Calculation:

Parameter p= true proportion of individuals in population

The best point of estimate of p is sample proportion

  X Is a random variable

  H0:p1=p2H1:p1>p2

  p^1=177434p^1=0.4078p^2=86222p^2=0.3874

The value of p is

  p^N(p,pqn)

  p Comes from null hypothesis and alternative hypothesis

  z=p^ppqn

Pooled proportion

  p=177+86434+222p=0.4009146

Standard of error

  SE=p(1p)(1n1+1n2)SE=0.4009146(10.4009146)(1434+1222)SE=0.0404

  z=p^1p^2SEz=0.40780.38740.0404z=0.5050

For two-tail test

  p Value

  =2P(z>z0)=2P(z>0.5050)=0.6136

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

Supplementing t-PA with endovascular therapy does not significantly increases the proportion of patients with an acute ischemic stroke

So, the regain functional independence.

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