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 26E
To determine

To find:

To perfom the test statistic for the following scenerio,

“A manufacturer of children’s vitamins claims that its vitamins are mixed so the each batch has exactly the following percentage of each color: 20% green, 40% yellow, 10% red, and 30% orange. To test the claim that these percentage are incorrect, 100 bottles of vitamins were pulled and the clors of the vitamins were tailled”.

Expert Solution & Answer
Check Mark

Answer to Problem 26E

Solution:

(a) H0:p1=0.02,p2=0.40,p3=0.10,andp4=0.30

Ha: There is some difference amongst the probabilities.

(b) Use 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=34.6485.

(d) There is sufficient evidence to support that percentages stated by the vitamin manufacturer are incorrect.

Explanation of Solution

The following table represents the random sample of number of vitamins and the color

of vitamins is,

Children's Vitamins
  Green Yellow Red orange
Number 1149 1948 552 1401

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 percentages stated by the vitamin manufacturer are not incorrect.

Null hypothesis: H0: The proportion of vitamins are not incorrect.

Alternative hypothesis: Ha: The proportion of vitamins are incorrect.

Let p1,p2,p3andp4 be the probabilities for Green, Red, Yellow and Orange respectively.

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

H0:p1=0.02,p2=0.40,p3=0.10,andp4=0.30

Ha: There is some difference amongst the probabilities.

Step 2: To determine whether the observed values proportion of color of vitamins is match with the expected values of proportion of color of vitamins. 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 of color of vitamins. Since we are assuming that the proportion percentages stated by the vitamin manufacturer are not incorrect is calculated as follows,

n=1149+1948++552+1401=5050

And,

E(x1)=np1=5050×0.20=1010E(x2)=np2=5050×0.40=2020E(x3)=np3=5050×0.10=505E(x4)=np4=5050×0.30=1515

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 5050.

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

Children's Vitamins Observed Values Expected Values OiEi  (OiEi)2  (OiEi)2Ei 
Green 1149 1010 139 19321 19.1297
Yellow 1948 2020 -72 5184 2.56634
Red 552 505 47 2209 4.37426
Orange 1401 1515 -114 12996 8.57822
          χ2=i=1n(OiEi)2Ei=34.6485

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 = 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=0.05 to find the critical value χ0.052 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,

χ0.052=7.815

The test statistic value is χ2=34.6485.

By using the above condition we reject the null hypothesis.

Since, 34.64857.815χ2χ0.052.

So, there is sufficient evidence to support that percentages stated by the vitamin manufacturer are incorrect.

Final statement:

(a) H0:p1=0.02,p2=0.40,p3=0.10,andp4=0.30

Ha: There is some difference amongst the probabilities.

(b) Use 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=34.6485.

(d) There is sufficient evidence to support that percentages stated by the vitamin manufacturer are incorrect.

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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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