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 11.2, Problem 38E

a)

To determine

To summarize data in a two-way table.

a)

Expert Solution
Check Mark

Explanation of Solution

Given:

    Group Treatment Number of patients Having stroke
    1 Placebo 1649 250
    2 Aspirin 1649 206
    3 Dipyridamole 1654 211
    4 Both 1650 157

Calculation:

For constructing a two-table, we need successes and failures with each treatment. Therefore, we already know about the treatment and number of patients had a stroke. So, number of patients had a no stroke will be calculated by subtracting number of patients had a stroke from number of patients. Following table shows the two-way table:

    Treatment Having stroke Having no stroke Total
    Placebo 250 1399 1649
    Aspirin 206 1443 1649
    Dipyridamole 211 1443 1654
    Both 157 1493 1650
    Total 824 5778 6602

b)

To determine

To perform chi square test of independence.

b)

Expert Solution
Check Mark

Answer to Problem 38E

There is convincing evidence that the difference in the effectiveness of the four treatments at the 0.05 level of significance.

Explanation of Solution

Given:

    Treatment Having stroke Having no stroke Total
    Placebo 250 1399 1649
    Aspirin 206 1443 1649
    Dipyridamole 211 1443 1654
    Both 157 1493 1650
    Total 824 5778 6602

Calculation:

Null and alternative hypotheses:

  H0 : The treatment and stroke are independent.

  Ha : The treatment and stroke are not independent.

Using excel formula,

        Having strock Having no stroke Total
    Placebo Observed 250 1399 1649
      Expected 205.81 1443.19 1649.00
      (O - E)² / E 9.49 1.35 10.84
    Aspirin Observed 206 1443 1649
      Expected 205.81 1443.19 1649.00
      (O - E)² / E 0.00 0.00 0.00
    Dipyridamole Observed 211 1443 1654
      Expected 206.44 1447.56 1654.00
      (O - E)² / E 0.10 0.01 0.12
    Both Observed 157 1493 1650
      Expected 205.94 1444.06 1650.00
      (O - E)² / E 11.63 1.66 13.29
    Total Observed 824 5778 6602
      Expected 824.00 5778.00 6602.00
      (O - E)² / E 21.22 3.03 24.24
    24.24 chi-square
    3 df
    2.22E-05 p-value

Decision: P-value = 2.22E-05< 0.05, reject H0.

Conclusion: There is convincing evidence that the difference in the effectiveness of the four treatments at the 0.05 level of significance.

Chapter 11 Solutions

PRACTICE OF STATISTICS F/AP EXAM

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