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Many track runners believe that they have a better chance of winning if they start in the inside lane that is closest to the field. For the data below, the lane closest to the field is Lane 1, the next lane is Lane 2, and so on until the outermost lane, Lane 6. The data lists the number of wins for track runners in the different starting positions. Calculate the chi-square test statistic to test the claim that the number of wins is uniformly distributed (equally likely) across the different starting positions. The results are based on 240 races.

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This question is incomplete, the complete question is;

Many track runners believe that they have a better chance of winning if they start in the inside lane that is closest to the field. For the data below, the lane closest to the field is Lane 1, the next lane is Lane 2, and so on until the outermost lane, Lane 6. The data lists the number of wins for track runners in the different starting positions. Calculate the chi-square test statistic to test the claim that the number of wins is uniformly distributed (equally likely) across the different starting positions. The results are based on 240 races.

starting position: 1 2 3 4 5 6

Number of wins : 44 33 50 32 45 36

options

a) 15.541

b) 9.326

c) 6.750

d) 12.592

Answer:

the chi-square test statistic is 6.75

Option c) 6.750 is the correct answer

Explanation:

Given that;

starting position: 1 2 3 4 5 6

Number of wins : 44 33 50 32 45 36

Given that we have 240 races

Using x² test.

we have to calculate the expected frequency

so

E(1) = 240 × 1/6 = 40

E(2) = 240 × 1/6 = 40

E(3) = 240 × 1/6 = 40

E(4) = 240 × 1/6 = 40

E(5) = 240 × 1/6 = 40

E(6) = 240 × 1/6 = 40

Now we know that,

x²_cal = [∑( Oi - Fi)²] / Fi

so

x²_cal = (44-40)²/40 + (33-40)²/40 + (50-40)²/40 + (32-40)²/40 + (45-40)²/40 + (35-40)²/40

= 0.4 + 1.225 + 2.5 + 1.6 + 0.625 + 0.4

= 6.75

Therefore the chi-square test statistic is 6.75

Option c) 6.750 is the correct answer

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