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 9.3, Problem 9.8E

i.

To determine

Critical value of x¯ used for rejecting H0 .

i.

Expert Solution
Check Mark

Answer to Problem 9.8E

Sincez(2.0400)>za(1.645)and a null hypothesis is rejected.

Explanation of Solution

Given:

  1. The size of the sample is n=35 .
  2. The mean x¯=2.4 .
  3. The standard deviation is s=0.29 .

Formula Used:

Test statistic:

  z=x¯μsn

Calculation:

Null hypothesis is H0:μ=2.3

Alternative hypothesis is H0:μ>2.3 (one-tailed)

So, for determining the tests statistics under the null hypothesis the test statistics is defined as below:

  z=x¯μs n =2.42.3 0.2 35 =2.0400

Thus, critical value as the significance level is α=0.05 and the test is one- tailed test, the rejection region is determined by a critical value with tail area equals to a α=0.05 .

So, H0 can be rejected if z>1.645 .

Conclusion:

Since z(2.0400)>za(1.645) and a null hypothesis is rejected.

ii.

To determine

  β=P

  (AcceptH0Whenμ=2.4)

ii.

Expert Solution
Check Mark

Answer to Problem 9.8E

  β=P is 0.3416 .

Explanation of Solution

Given:

  μ=2.4

Formula Used:

Test statistic:

  z=x¯μsn

Calculation:

The critical value at 5% level of significance is 1.645 .

  z=x¯μsn>1.645

So, it implies that the acceptance region is,

  x¯2.4 0.29 35 <1.645x¯<2.38β=P(AcceptH0Whenμ=2.4)=P( x¯<2.38/μ=2.4)=P(z= x ¯ μ s n < 2.382.4 0.29 35 )=P(z<0.4080)=0.3416

The power of the test is,

  1β=10.3416=0.6584

Conclusion:

Thus, β=P=0.3416 .

iii.

To determine

  β when μ=2.3,2.5,2.6

iii.

Expert Solution
Check Mark

Answer to Problem 9.8E

  β when μ=2.3 is 0.9487 , β when μ=2.5 is 0.007 , β when μ=2.6 is 0.000004

Explanation of Solution

Given:

  μ=2.3,2.5,2.6

Formula Used:

Test statistic:

  z=x¯μsn

Calculation:

Initially, to find out β (AcceptH0Whenμ=2.3)

The critical value at 5% level of significance is 1.645 .

  z=x¯μsn>1.645

So, it implies that the acceptance region is,

  x¯2.3 0.29 35 <1.645x¯<2.38β=P(AcceptH0Whenμ=2.3)=P( x¯<2.38/μ=2.3)=P(z= x ¯ μ s n < 2.382.3 0.29 35 )=P(z<1.6320)=0.9487

The power of the test is,

  1β=10.9487=0.0513

Now, to find out β (AcceptH0Whenμ=2.5)

The critical value at 5% level of significance is 1.645 .

  z=x¯μsn>1.645

So, it implies that the acceptance region is,

  x¯2.5 0.29 35 <1.645x¯<2.38β=P(AcceptH0Whenμ=2.5)=P( x¯<2.38/μ=2.5)=P(z= x ¯ μ s n < 2.382.5 0.29 35 )=P(z<2.4480)=0.007

The power of the test is,

  1β=10.007=0.9928

So, to find out β (AcceptH0Whenμ=2.6)

The critical value at 5% level of significance is 1.645 .

  z=x¯μsn>1.645

So, it implies that the acceptance region is,

  x¯2.6 0.29 35 <1.645x¯<2.38β=P(AcceptH0Whenμ=2.6)=P( x¯<2.38/μ=2.6)=P(z= x ¯ μ s n < 2.382.6 0.29 35 )=P(z<4.48806)=0.000004

The power of the test is,

  1β=10.000004=0.9999

Conclusion:

Therefore, β when μ=2.3 is 0.9487 , β when μ=2.5 is 0.007 , β when μ=2.6 is 0.000004

iv.

To determine

To draw: Graph the power curve for the test using values of β from part (ii) and part (iii).

iv.

Expert Solution
Check Mark

Answer to Problem 9.8E

Power curve for the test is shown as below:

  Introduction to Probability and Statistics, Chapter 9.3, Problem 9.8E , additional homework tip  1

Explanation of Solution

Given:

Values of β ,

  1. β=0.3416
  2. β=0.9487
  3. β=0.007
  4. β=0.000004

Calculation:

By considering the following table,

    True mean (μ)Power of the test(1β)
    2.30.0513
    2.40.6584
    2.50.9928
    2.60.9999

Power curve for the test is shown as below:

  Introduction to Probability and Statistics, Chapter 9.3, Problem 9.8E , additional homework tip  2

Conclusion:

Therefore, Power curve for the test is shown as below:

  Introduction to Probability and Statistics, Chapter 9.3, Problem 9.8E , additional homework tip  3

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Introduction to Probability and Statistics

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