Question

The safety director of a large steel mill took samples at random from company records of minor work-related accidents and classified them according to the time the accident took place.

	Number of					Number of
Time	Accidents				Time	Accidents
8 up to 9 a.m.		10			1 up to 2 p.m.		8
9 up to 10 a.m.		7			2 up to 3 p.m.		8
10 up to 11 a.m.		8			3 up to 4 p.m.		6
11 up to 12 p.m.		22			4 up to 5 p.m.		17

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Using the goodness-of-fit test and the 0.01 level of significance, determine whether the accidents are evenly distributed throughout the day.

H₀: The accidents are evenly distributed throughout the day.   
H₁: The accidents are not evenly distributed throughout the day.

State the decision rule, using the 0.01 significance level. (Round your answer to 3 decimal places.)

Compute the value of chi-square. (Round your answer to 3 decimal places.)

What is your decision regarding H₀?

Answer 1

Given table data is as below MATRIX col1 col2 TOTALS row 1 10 8 18 row 2 7 8 15 row 3 8 6 14 row 4 22 17 39 TOTALS 47 39 86 ------------------------------------------------------------------ calculation formula for E table matrix E-TABLE col1 col2 row 1 row1*col1/N row1*col2/N row 2 row2*col1/N row2*col2/N row 3 row3*col1/N row3*col2/N row 4 row4*col1/N row4*col2/N ------------------------------------------------------------------ expected frequecies calculated by applying E - table matrix formulae E-TABLE col1 col2 row 1 9.837 8.163 row 2 8.198 6.802 row 3 7.651 6.349 row 4 21.314 17.686 ------------------------------------------------------------------ calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 10 9.837 0.163 0.027 0.003 8 8.163 -0.163 0.027 0.003 7 8.198 -1.198 1.435 0.175 8 6.802 1.198 1.435 0.211 8 7.651 0.349 0.122 0.016 6 6.349 -0.349 0.122 0.019 22 21.314 0.686 0.471 0.022 17 17.686 -0.686 0.471 0.027 ᴪ^2 o = 0.476 ------------------------------------------------------------------ set up null vs alternative as null, Ho: no relation b/w X and Y OR X and Y are independent alternative, H1: exists a relation b/w X and Y OR X and Y are dependent level of significance, α = 0.01 from standard normal table, chi square value at right tailed, ᴪ^2 α/2 =11.345 since our test is right tailed,reject Ho when ᴪ^2 o > 11.345 we use test statistic ᴪ^2 o = Σ(Oi-Ei)^2/Ei from the table , ᴪ^2 o = 0.476 critical value the value of |ᴪ^2 α| at los 0.01 with d.f (r-1)(c-1)= ( 4 -1 ) * ( 2 - 1 ) = 3 * 1 = 3 is 11.345 we got | ᴪ^2| =0.476 & | ᴪ^2 α | =11.345 make decision hence value of | ᴪ^2 o | < | ᴪ^2 α | and here we do not reject Ho ᴪ^2 p_value =0.924 ANSWERS --------------- null, Ho: no relation b/w X and Y OR X and Y are independent alternative, H1: exists a relation b/w X and Y OR X and Y are dependent test statistic: 0.476 critical value: 11.345 p-value:0.924 decision: do not reject Ho

we do not have enough evidence to support the claim that   whether the accidents are evenly distributed throughout the day.