Question

1. Perform the indicated goodness-of-fit test.

Use a significance level of 0.01 to test the claim that workplace accidents are distributed on workdays as follows: Monday 25%, Tuesday: 15%, Wednesday: 15%, Thursday: 15%, and Friday: 30%.
In a study of 100 workplace accidents, 26 occurred on a Monday, 15 occurred on a Tuesday, 17 occurred on a Wednesday, 17 occurred on a Thursday, and 25 occurred on a Friday.

a. test statistic.

b. critical value

2. Perform the indicated goodness-of-fit test.

You roll a die 48 times with the following results.

a. test statistic

b. critical value
Use a significance level of 0.05 to test the claim that the die is fair.

Answer 1

Observed (Oi )	Expected ( Ei)	Oi-Ei	(Oi-Ei)^2	(Oi-Ei)^2/Ei
0.26	0.25	0.01	0.0001	0.0004
0.15	0.15	0	0	0
0.17	0.15	0.02	0.0004	0.0027
0.17	0.15	0.02	0.0004	0.0027
0.25	0.3	-0.05	0.0025	0.0083

null hypothesis: true proportions in all 5 categories is equal to their expected value alternative hypothesis: atleast one proportions is different from its expected value. HO: pl = 0.25, P2 = 0.15, p3 = 0.15, P4 = 0.15, p5 = 0.3 H1: atleast one p do not match the claimed value level of significance, alpha = 0.01 from standard normal table, chi square value at right tailed, y^2 alpha/2 =13.2767 since our test is right tailed,reject Ho when y^2 0 > 13.2767 we use test statistic y^2 0 = E(Oi-Ei)^2/Ei from the table take sum of (Oi-Ei)^2/Ei, we get y^2 0 = 0.0141 critical value the value of Y^2 alpha at los 0.01 with d.f, n - 1 = 5 - 1 = 4 is 13.2767 we got | ^2 =0.0141 & X^2 alpha = 13.277 make decision hence value of 4^2 0 | < ^2 alpha and here we do not reject Ho since test statistic is not greater than critical value, we (fail to reject null hypothesis. at 1%. level, there is not significant evidence that atleast one proportions is different from its expected value. Y^2 0 = 0.0141 critical value = 13.277