Use the following contingency table to complete? (a) and? (b)below.
| | A | | B | | C | | Total | |
1 | | 15 | | 30 | | 45 | | 90 | | |
2 | | 45 | | 50 | | 55 | | 150 | | |
Total | | 60 | | 80 | | 100 | | 240 | | |
a. Compute the expected frequencies for each cell.
| | A | | B | | C | |
1 | | | | | | | |
2 | | | | | | | |
| | ?(Type integers or? decimals.) |
b. Compute ?2STAT. is it significant at ?=0.005??
Set up the null and alternative hypotheses to test. Choose thecorrect answer below.
- H0?: ?A= ?B= ?C
H1?:Not all ?j are equal? (where j=?A, ?B, C)
- H0?: ?1=?2
H1?: Not all j?j are equal? (where j=?1, ?2)
- H0?: Not all j?j are equal? (where j=?A, ?B, C)
H1?: ?A= ?B= ?C
- H0?: Not all j?j are equal? (where j=?1, ?2)
H1?: ?1=?2
Compute ?2STAT.
?2STAT=
?(Round to three decimal places as? needed.)
Find the? p-value.
?p-value=
?(Round to three decimal places as? needed.)
Is ?2STAT significant at ?=0.005??
- Yes?, because the? p-value is greater than or equal to thelevel of significance.
- Yes?, because the? p-value is less than the level ofsignificance.
- No?, because the? p-value is less than the level ofsignificance.
- No because the? p-value is greater than or equal to the levelof significance.
