Beginning Statistics, 2nd Edition
Beginning Statistics, 2nd Edition
2nd Edition
ISBN: 9781932628678
Author: Carolyn Warren; Kimberly Denley; Emily Atchley
Publisher: Hawkes Learning Systems
Question
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Chapter 10.6, Problem 21E
To determine

To find:

To perfom the test statistic for the following scenerio,

“A school principal claims that the number of students who are tardy in her school does not vary from month to month. Using a 0.05 level of significance, test a teacher’s claim that the number of tardy studens does vary by the month”.

Expert Solution & Answer
Check Mark

Answer to Problem 21E

Solution:

(a) H0:p1=p2=...=p10

Ha: There is some difference amongst the probabilities.

(b) The chi-square test for this goodness of fit and the level of siginificance is α=0.05.

(c) The test statistic value is χ2=15.56.

(d) There is no sufficient evidence to support the teacher’s claim that the number of tardy students does vary by the month.

Explanation of Solution

The following table represents the random sample of number of students who are tardy in her school is,

Tardy Students
Aug. Sept. Oct. Nov. Dec. Jan. Feb. Mar. Apr. May
Number 7 18 16 5 8 12 15 18 11 15

Step 1: To state the null and alternative hypotheses.

When starting the hypotheses to be used we take the null hypothesis to be that the proportion of students who are tardy in her school does not vary from month to month.

Null hypothesis: H0: The proportion of of students who are tardy in her school does not vary from month to month.

Alternative hypothesis: Ha: The proportion of of students who are tardy in her school does vary from month to month.

Mathematically, we can write the null and alternative hypothesis as follows,

H0:p1=p2=...=p10

Ha: There is some difference amongst the probabilities.

Step 2: To determine whether the observed values of tardy students match with the expected values of tardy students. Since we assume that the necessary condition that have been met. So, use the chi-square test for this goodness of fit and the level of siginificance is α=0.05.

Step 3: To calculate the test statistic.

Let‘s calculate the expected value for each month of the year. Since we are assuming that the number of tardy students does not vary for each month, then the probability will be same for every day, so expected number of students for each month is calculated as follows,

n=7+18+16+5+8+12+15+18+11+15=125

And,

E(xi)=npi=125.110=12.5

Formula for calculating test statistic for a Chi-Square test for goodness of fit:

The test statistic for a chi-square test for goodness of fit is given below,

χ2=i=1n(OiEi)2Ei

Where Oi, is the observed frequency for the ith possible outcome and Ei is the expected frequency for the ith possible outcome and n is the sample size.

From the above table represent the value of n is 125.

The following table represents the test statistic, χ2, for a chi- square test for goodness of fit is given below,

Month Observed Values Expected values OiEi  (OiEi)2  (OiEi)2Ei 
Aug. 7 12.5 -5.5 30.25 2.42
Sept. 18 12.5 5.5 30.25 2.42
Oct. 16 12.5 3.5 12.25 0.98
Nov. 5 12.5 -7.5 56.25 4.5
Dec. 8 12.5 -4.5 20.25 1.62
Jan. 12 12.5 -0.5 0.25 0.02
Feb. 15 12.5 2.5 6.25 0.5
Mar. 18 12.5 5.5 30.25 2.42
Apr. 11 12.5 -1.5 2.25 0.18
May 15 12.5 2.5 6.25 0.5
          χ2=i=1n(OiEi)2Ei=15.56

Step 4: Degrees of freedom in a Chi-square test for goodness of fit:

In a chi-square test for goodness of fit the number of degrees of freedom for the chi-square distribution of the test statistic is given by,

df=k1

Where k is the number of possible outcomes for each trial.

Rejection Region for Chi-Square Test for Association:

Reject the null hypothesis, H0, if χ2χα2.

From the given information k = 10.

Substitute the above values in the formula of degrees of freedom to get the following,

df=k1=101=9

Conclusion:

Use the level of significance of α=0.05 to find the critical value χ0.052 of the chi-square distribution with df = 9 in order to make a decision about the null hypothesis.

Use the “Area to the right of the critical value χ2” table to get the following,

χ0.052=16.9190

The test statistic value is χ2=15.56.

By using the above condition we fail to reject the null hypothesis.

Since 15.56<16.9190χ2<χ0.052.

So, there is no sufficient evidence to support the teacher’s claim that the number of tardy students does vary by the month.

Final statement:

(a) H0:p1=p2=...=p10

Ha: There is some difference amongst the probabilities.

(b) The chi-square test for this goodness of fit and the level of siginificance is α=0.05.

(c) The test statistic value is χ2=15.56.

(d) There is no sufficient evidence to support the teacher’s claim that the number of tardy students does vary by the month.

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Chapter 10 Solutions

Beginning Statistics, 2nd Edition

Ch. 10.1 - Prob. 11ECh. 10.1 - Prob. 12ECh. 10.1 - Prob. 13ECh. 10.1 - Prob. 14ECh. 10.1 - Prob. 15ECh. 10.1 - Prob. 16ECh. 10.1 - Prob. 17ECh. 10.1 - Prob. 18ECh. 10.1 - Prob. 19ECh. 10.1 - Prob. 20ECh. 10.1 - Prob. 21ECh. 10.1 - Prob. 22ECh. 10.1 - Prob. 23ECh. 10.1 - Prob. 24ECh. 10.1 - Prob. 25ECh. 10.1 - Prob. 26ECh. 10.1 - Prob. 27ECh. 10.1 - Prob. 28ECh. 10.1 - Prob. 29ECh. 10.1 - Prob. 30ECh. 10.2 - Prob. 1ECh. 10.2 - Prob. 2ECh. 10.2 - Prob. 3ECh. 10.2 - Prob. 4ECh. 10.2 - Prob. 5ECh. 10.2 - Prob. 6ECh. 10.2 - Prob. 7ECh. 10.2 - Prob. 8ECh. 10.2 - Prob. 9ECh. 10.2 - Prob. 10ECh. 10.2 - Prob. 11ECh. 10.2 - Prob. 12ECh. 10.2 - Prob. 13ECh. 10.2 - Prob. 14ECh. 10.2 - Prob. 15ECh. 10.2 - Prob. 16ECh. 10.2 - Prob. 17ECh. 10.2 - Prob. 18ECh. 10.2 - Prob. 19ECh. 10.2 - Prob. 20ECh. 10.2 - Prob. 21ECh. 10.2 - Prob. 22ECh. 10.2 - Prob. 23ECh. 10.2 - Prob. 24ECh. 10.2 - Prob. 25ECh. 10.2 - Prob. 26ECh. 10.2 - Prob. 27ECh. 10.2 - Prob. 28ECh. 10.2 - Prob. 29ECh. 10.2 - Prob. 30ECh. 10.3 - Prob. 1ECh. 10.3 - Prob. 2ECh. 10.3 - Prob. 3ECh. 10.3 - Prob. 4ECh. 10.3 - Prob. 5ECh. 10.3 - Prob. 6ECh. 10.3 - Prob. 7ECh. 10.3 - Prob. 8ECh. 10.3 - Prob. 9ECh. 10.3 - Prob. 10ECh. 10.3 - Prob. 11ECh. 10.3 - Prob. 12ECh. 10.3 - Prob. 13ECh. 10.3 - Prob. 14ECh. 10.3 - Prob. 15ECh. 10.3 - Prob. 16ECh. 10.3 - Prob. 17ECh. 10.3 - Prob. 18ECh. 10.3 - Prob. 19ECh. 10.4 - Prob. 1ECh. 10.4 - Prob. 2ECh. 10.4 - Prob. 3ECh. 10.4 - Prob. 4ECh. 10.4 - Prob. 5ECh. 10.4 - Prob. 6ECh. 10.4 - Prob. 7ECh. 10.4 - Prob. 8ECh. 10.4 - Prob. 9ECh. 10.4 - Prob. 10ECh. 10.5 - Prob. 1ECh. 10.5 - Prob. 2ECh. 10.5 - Prob. 3ECh. 10.5 - Prob. 4ECh. 10.5 - Prob. 5ECh. 10.5 - Prob. 6ECh. 10.5 - Prob. 7ECh. 10.5 - Prob. 8ECh. 10.5 - Prob. 9ECh. 10.5 - Prob. 10ECh. 10.5 - Prob. 11ECh. 10.5 - Prob. 12ECh. 10.5 - Prob. 13ECh. 10.5 - Prob. 14ECh. 10.5 - Prob. 15ECh. 10.6 - Prob. 1ECh. 10.6 - Prob. 2ECh. 10.6 - Prob. 3ECh. 10.6 - Prob. 4ECh. 10.6 - Prob. 5ECh. 10.6 - Prob. 6ECh. 10.6 - Prob. 7ECh. 10.6 - Prob. 8ECh. 10.6 - Prob. 9ECh. 10.6 - Prob. 10ECh. 10.6 - Prob. 11ECh. 10.6 - Prob. 12ECh. 10.6 - Prob. 13ECh. 10.6 - Prob. 14ECh. 10.6 - Prob. 15ECh. 10.6 - Prob. 16ECh. 10.6 - Prob. 17ECh. 10.6 - Prob. 18ECh. 10.6 - Prob. 19ECh. 10.6 - Prob. 20ECh. 10.6 - Prob. 21ECh. 10.6 - Prob. 22ECh. 10.6 - Prob. 23ECh. 10.6 - Prob. 24ECh. 10.6 - Prob. 25ECh. 10.6 - Prob. 26ECh. 10.7 - Prob. 1ECh. 10.7 - Prob. 2ECh. 10.7 - Prob. 3ECh. 10.7 - Prob. 4ECh. 10.7 - Prob. 5ECh. 10.7 - Prob. 6ECh. 10.7 - Prob. 7ECh. 10.7 - Prob. 8ECh. 10.7 - Prob. 9ECh. 10.7 - Prob. 10ECh. 10.7 - Prob. 11ECh. 10.7 - Prob. 12ECh. 10.7 - Prob. 13ECh. 10.7 - Prob. 14ECh. 10.7 - Prob. 15ECh. 10.7 - Prob. 16ECh. 10.7 - Prob. 17ECh. 10.7 - Prob. 18ECh. 10.7 - Prob. 19ECh. 10.7 - Prob. 20ECh. 10.CR - Prob. 1CRCh. 10.CR - Prob. 2CRCh. 10.CR - Prob. 3CRCh. 10.CR - Prob. 4CRCh. 10.CR - Prob. 5CRCh. 10.CR - Prob. 6CRCh. 10.CR - Prob. 7CRCh. 10.CR - Prob. 8CRCh. 10.CR - Prob. 9CRCh. 10.CR - Prob. 10CRCh. 10.CR - Prob. 11CRCh. 10.CR - Prob. 12CRCh. 10.CR - Prob. 13CRCh. 10.CR - Prob. 14CRCh. 10.CR - Prob. 15CRCh. 10.P - Prob. 1PCh. 10.P - Prob. 2PCh. 10.P - Prob. 3PCh. 10.P - Prob. 4PCh. 10.P - Prob. 5PCh. 10.P - Prob. 6PCh. 10.P - Prob. 7PCh. 10.P - Prob. 8PCh. 10.P - Prob. 9PCh. 10.P - Prob. 10P
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