Consider the contingency table presented below. A hypothesis test for independen
ID: 3219068 • Letter: C
Question
Consider the contingency table presented below. A hypothesis test for independence is conducted to determine if the number of nonconforming parts produced is independent of time of day.
Time
Conforming Parts
Nonconforming Parts
Totals
Day
94
5
99
Evening
87
10
97
Overnight
65
23
88
Totals
246
38
284
Answer the following questions:
What type of test should be done? _____________________________
What is the calculated test statistic equal to? _____________________________
What is the critical value equal to? _____________________________
Should we reject or fail to reject the null hypothesis? _____________________________
Time
Conforming Parts
Nonconforming Parts
Totals
Day
94
5
99
Evening
87
10
97
Overnight
65
23
88
Totals
246
38
284
Explanation / Answer
1)What type of test should be done? chi square test of independence
calculated test statistic =19.071
degree of freedom =(r-1)*(c-1)= 2
as level of significance not given we will consider it 0.05.
critical value =5.9915
as test stat is higher then critical value we will reject null hypothesis.
Observed O conforming Non conforming Total day 94 5 99 evening 87 10 97 Overnight 65 23 88 Total 246 38 284 Expected E=rowtotal*column total/grand total conforming Non conforming Total day 85.754 13.246 99 evening 84.021 12.979 97 Overnight 76.225 11.775 88 Total 246 38 284 chi square =(O-E)^2/E conforming Non conforming Total day 0.793 5.134 5.927 evening 0.106 0.684 0.789 Overnight 1.653 10.702 12.355 Total 2.552 16.519 19.071