Introduction to Probability and Statistics
Introduction to Probability and Statistics
14th Edition
ISBN: 9781133103752
Author: Mendenhall, William
Publisher: Cengage Learning
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Chapter 14, Problem 14.48SE
To determine

To identify:

The value of p and degrees of freedom of x2 .

Expert Solution & Answer
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Answer to Problem 14.48SE

The value of P is 0.04 and null hypothesis is rejected and the data provide sufficient evidence to indicate there is a significant difference between actual and experimental observations.

Explanation of Solution

Given information:

The problem experience consists of 4 tails was repeated 100 times.

The number of success gets as:

    No of successesNo of times obtained

      0


      11

      1


      17

      2


      42

      3


      21

      4


      9

Calculation:

As know that,

  p=x¯n

Where,

  n= number of trials.

  n=4

Now, for estimating p ,

So, consider

  x¯= x i f i f i x¯=200100x¯=2

Put the value of (x¯=2,n=4) in the equation of p

  p=x¯np=24p=0.5

Now, the expected values for a binomial experiment of 4 trails are,

  Exp=N×(nC xipxiqnxi)

Are 0.0625,0.25,0.375,0.25,0.0625 respectively.

Now, for testing whether the data provide sufficient evidence to indicate there is any significance difference between actual and the experimental observations or not.

The null hypothesis is,

  H0: There is sufficient evidence to indicate there is a significance difference between actual observation and experimental observations.

The alternative hypothesis is,

  H1: There is sufficient evidence to indicate there is a significance difference between actual observation and experimental observations.

In case test statistic x2 is too large, so there is a rejection of null hypothesis, H0 .

    Obs. countExp. countObs.-exp

      (Obsexp)2


      ( Obsexp)2exp

      11


      6.25

      4.75

      22.5625

      3.61

      17


      25

      8

      64

      2.56

      42


      37.5

      4.5

      20.25

      0.54

      21


      25

      4

      16

      0.64

      9


      6.25

      2.75

      7.5625

      1.21

The chi-square statistic is,

  x2= ( Obsexp ) 2 exp=8.55558.56

Test statistic has (k11) degrees of freedom,

  df=(k11)df=(k2)df=(52)df=3

The P-value can be determined as,

  P(x2>8.56)=1P(x2>8.56)P(x2>8.56)=10.96P(x2>8.56)=0.04

Thus, the value of P is 0.04 which is smaller than the 0.05=α .

So, there is rejection of the null hypothesis because of low p value.

Thus, the data provide sufficient evidence to indicate there is a significant difference between actual and experimental observations.

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