Test Statistic: The general form of the test statistic for goodness-of-fit or an
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Question
Test Statistic:
The general form of the test statistic for goodness-of-fit or analysis of contingency tables is given by
where O is the observed frequency in a cell and E is the expected frequency in a cell. The null hypothesis is rejected only if the test statistic is large. That is, the test is right-tailed.
Problem 1. A hole was drilled in a die and was filled with a lead weight. Then the die was rolled 200 times. Here are the observed frequencies of the outcomes of 1, 2, 3, 4, 5, and 6 respectively:
27, 31, 42, 40, 28, 32.
Use a 0.05 significance level to test the claim that the outcomes are not equally likely.
Let pi be the probability that the ith face of the die appears when it is rolled, i=1,2,3,4,5,6. Write down the null and alternate hypothesis of the test.
What is the value of the chi-square test statistic?
What is the degrees of freedom of the test statistic under the null hypothesis?
What is the p-value of the test? Do you reject the null hypothesis? Why?
What is your conclusion based on (4)?
Problem 2. Refer to the following contingency table. Do the data present sufficient evidence to indicate that the type of furniture defect varies with the shift during which the piece of furniture is produced?
Shift
Type of Defects
1
2
3
A
15
26
33
B
21
31
17
C
45
34
49
D
13
5
20
Write down the null and alternate hypothesis of the test.
What is observed value of chi-square test statistic?
What is the degrees of freedom of the test statistic under the null hypothesis?
What is the p-value of the test? Do you reject the null hypothesis at significance level 0.01? Why?
What is your conclusion based on (4)?
Shift
Type of Defects
1
2
3
A
15
26
33
B
21
31
17
C
45
34
49
D
13
5
20
Explanation / Answer
Solution:-
2)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
H0: The type of furniture defect does not varies with the shift during which the piece of furniture is produced.
Ha: The type of furniture defect varies with the shift during which the piece of furniture is produced.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square test for independence.
Analyze sample data. Applying the chi-square test for independence to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.
DF = (r - 1) * (c - 1) = (4 - 1) * (3 - 1)
D.F = 6
Er,c = (nr * nc) / n
2 = [ (Or,c - Er,c)2 / Er,c ]
2 = 25.20 + 30.83 + 21.93
2 = 77.96
where DF is the degrees of freedom.
The P-value is the probability that a chi-square statistic having 6 degrees of freedom is more extreme than 77.96.
We use the Chi-Square Distribution Calculator to find P(2 > 77.96) = less than 0.00001
Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we cannot accept the null hypothesis.
From the above test we have sufficient evidence to indicate that the type of furniture defect varies with the shift during which the piece of furniture is produced.
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