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.P, Problem 10P
To determine

To find:

To write is there sufficient evidence to conclude at α=0.10 that the makeup of your institution has significantly changed between Year 1 and Year 2.

Expert Solution & Answer
Check Mark

Answer to Problem 10P

Solution:

There is no sufficient evidence to conclude that the makeup of your institution has significantly changed between Year 1 and Year 2.

Explanation of Solution

Consider two years that are not consecutive from which to collect data.

That is choosing the current academic year for Year 2 and the academic year four years earlier for Year 1.

Suppose the 2015 for year 1 and 2019 for year 2.

For year 1:

The number of students who were enrolled at your institution for each classification during year 1 is given below,

  2015
Number of Freshmen Enrolled 15
Number of Sophomores Enrolled 18
Number of Juniors Enrolled 21
Number of Seniors Enrolled 73
Total Number of Students Enrolled 127

Let p1,p2,p3andp4 be the probabilities for freshmen, sophomores, junior and senior respectively.

Mathematically, we can write the null and alternative hypothesis for four different students as follows,

H0:p1=0.1181,p2=0.1417,p3=0.1654,andp4=0.5748

Ha: There is some difference amongst the probabilities.

Let‘s calculate the expected value for year 1. Since we are assuming that the proportion percentages stated for the student in year 1 are not incorrect is calculated as follows,

n=15+18+21+73=127

And,

E(x1)=np1=127×0.1181=14.9987E(x2)=np2=127×0.1417=17.9959E(x3)=np3=127×0.1654=21.0058E(x4)=np4=127×0.5748=72.9996

For year 2:

The number of students who were enrolled at your institution for each classification during year 2 is given below,

  2015
Number of Freshmen Enrolled 24
Number of Sophomores Enrolled 20
Number of Juniors Enrolled 19
Number of Seniors Enrolled 20

Let p1,p2,p3andp4 be the probabilities for freshmen, sophomores, junior and senior respectively.

Mathematically, we can write the null and alternative hypothesis for four different students as follows,

  2019
Percentage of Freshmen Enrolled 28.92%
Percentage of Sophomores Enrolled 24.10%
Percentage of Juniors Enrolled 22.89%
Percentage of Seniors Enrolled 24.10%

H0:p1=0.2892,p2=0.2410,p3=0.2289,andp4=0.2410

Ha: There is some difference amongst the probabilities.

Let‘s calculate the expected value for year 2. Since we are assuming that the proportion percentages stated for the student in year 2 are not incorrect is calculated as follows,

n=24+20+19+20=83

And,

E(x1)=np1=83×0.2892=24.0036E(x2)=np2=83×0.2410=20.003E(x3)=np3=83×0.2289=18.9987E(x4)=np4=83×0.2410=20.003

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 for year 1 is 127 and for year 2 is 83.

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

Year 1
Category Observed Values Expected values (OiEi) (OiEi)2 (OiEi)2Ei
Freshman 15 14.9987 0.0013 1.69×106 1.1268×107
Sophomore 18 17.9959 0.0041 1.681×105 9.341×107
Junior 21 21.0058 -0.0058 3.364×105 1.6015×106
Senior 73 72.9996 0.0004 1.6×107 2.1918×109
χ2=i=1n(OiEi)2Ei=2.6504×106
Year 2
Freshman 24 24.0036 -0.0036 1.296×105 5.39919×107
Sophomore 20 20.003 -0.003 9×106 4.49933×107
Junior 19 18.9987 0.0013 1.69×106 8.89535×108
Senior 20 20.003 -0.003 9×106 4.49933×107
χ2=i=1n(OiEi)2Ei=1.52874×106

The calculated test statistic for year 1 and for year 2 is 2.6504×106 and 1.52874×106 respectively.

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 = 4.

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

df=k1=41=3

Conclusion:

Use the level of significance of α=0.10 to find the critical value χ0.102 of the chi-square distribution with df = 3 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,

χ0.102=6.2514

Year 1:

The test statistic value is χ2=2.6504×106.

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

Since 2.6504×106<6.2514χ2<χ0.102.

Year 2:

The test statistic value is χ2=1.52874×106.

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

Since 1.52874×106<6.2514χ2<χ0.102.

The null hypothesis will be accepted for both year 1 and year 2.

Hence the proportion percentage for the students does not vary for both year 1 and year 2.

So, there is no sufficient evidence to conclude that the makeup of your institution has significantly changed between Year 1 and Year 2. Since the proportion percentage will not vary between two years. That is, percentage of students in every classification in year 1will be equal to percentage of students in every classification in year 2.

Final statement:

There is no sufficient evidence to conclude that the makeup of your institution has significantly changed between Year 1 and Year 2.

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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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