Statistics, Binder Ready Version: Unlocking the Power of Data
Statistics, Binder Ready Version: Unlocking the Power of Data
2nd Edition
ISBN: 9781119163664
Author: Robin H. Lock, Patti Frazer Lock, Kari Lock Morgan, Eric F. Lock, Dennis F. Lock
Publisher: WILEY
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Chapter D, Problem 1E
To determine

Test whether the data provide that the numbers of bills are not equally valued among three different servers.

Expert Solution & Answer
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Answer to Problem 1E

There is sufficient evidence to conclude that the numbers of bills are not equally valued among three different servers.

Explanation of Solution

Denote p1,p2, and p3 be the proportions for A, B, and C, respectively.

The null and alternative hypotheses can be written as follows:

Null hypothesis:

H0:p1=p2=p3=13.

Alternative hypothesis:

Ha:At least one pi(i=1,2,3) is not specified as in H0.

The total observed counts for A, B, and C are 60, 65, and 32, respectively.

By adding all the observed counts, the total numbers of observations are 157.

If all the proportions are equally likely, the expected count for each answer is given below:

E(pi)=npi=157(13)52.3

The χ2test statistic is calculated as follows:

χ2=(ObservedExpected)2Expected=[(6052.3)252.3+(6552.3)252.3+(3252.3)252.3]=1.134+3.084+7.87912.09

There are 3 attributes in the test. Therefore, the degrees of freedom for the Chi-square distribution for the table are given below:

df=k1=31=2

Step-by-step procedure to obtain the p-value using StatKey software:

  • Click on χ2, enter degrees of freedom as 2.
  • Click Ok.
  • Select right tail at the top-left of the screen.
  • Double click on the cutoff value on the X-axis.
  • Enter cutoff value as 12.09.
  • Click Ok.

Output obtained using StatKey software is given below:

Statistics, Binder Ready Version: Unlocking the Power of Data, Chapter D, Problem 1E

From the above result, the p-value for the test is 0.002.

Decision rule:

If the p-value is less than or equal to the level of significance, reject the null hypothesis. Otherwise, do not reject the null hypothesis.

Conclusion:

Here, the p-value is 0.002.

Therefore, the p-value is less than any common levels of significance such as 0.10, 0.05, and 0.01.

Hence, reject the null hypothesis.

Therefore, there is sufficient evidence to conclude that the numbers of bills are not equally valued among three different servers.

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