(a) The chi-square test of independence is appropriate because we want to determine if there is a significant relationship between two nominal (categorical) variables.
(b) The hypothesis being tested is:
H0: A character dying on the show and uniform color are independent
Ha: A character dying on the show and uniform color are not independent
(c)
Col 1 | Col 2 | Total | ||
Row 1 | Observed | 129 | 7 | 136 |
Expected | 123.35 | 12.65 | 136.00 | |
O - E | 5.65 | -5.65 | 0.00 | |
(O - E)² / E | 0.26 | 2.52 | 2.78 | |
Row 2 | Observed | 46 | 9 | 55 |
Expected | 49.88 | 5.12 | 55.00 | |
O - E | -3.88 | 3.88 | 0.00 | |
(O - E)² / E | 0.30 | 2.95 | 3.25 | |
Row 3 | Observed | 215 | 24 | 239 |
Expected | 216.77 | 22.23 | 239.00 | |
O - E | -1.77 | 1.77 | 0.00 | |
(O - E)² / E | 0.01 | 0.14 | 0.15 | |
Total | Observed | 390 | 40 | 430 |
Expected | 390.00 | 40.00 | 430.00 | |
O - E | 0.00 | 0.00 | 0.00 | |
(O - E)² / E | 0.58 | 5.61 | 6.19 | |
6.19 | chi-square | |||
2 | df | |||
.0453 | p-value |
(d) The test statistic is 6.19. The degrees of freedom is 2.
(e) The p-value is 0.0453.
Since the p-value (0.0453) is less than the significance level (0.05), we can reject the null hypothesis.
Therefore, we have sufficient evidence to conclude that a character dying on the show and uniform color are not independent.
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