Question

A bus company claims that 70% of its services are less than two minutes late, 20% are between two and five minutes late, and only 10% are more than five minutes late.

A consumers' group records the times that 120 randomly chosen services are late at arbitrary points along their routes, and finds that 79 of the services are less than two minutes late, 25 of the services are between two and five minutes late, and 16 are more than five minutes late.

The consumers' group performs a Chi-squared goodness-of-fit test of (H₀: the bus company's percentages are correct) against (H₁: the bus company's percentages are incorrect).

What is the numerical value of the test statistic?

Answer 1

Category	Observed Frequency (O)	Proportion, p	Expected Frequency (E)	(O-E)²/E
Less than 2 min	79	0.70	120 * 0.7 = 84	(79 - 84)²/84 = 0.2976
2 to 5 min	25	0.20	120 * 0.2 = 24	(25 - 24)²/24 = 0.0417
More than 5 min	16	0.10	120 * 0.1 = 12	(16 - 12)²/12 = 1.3333
Total	120	1.00	120	1.6726

Null and Alternative hypothesis:

Ho: The bus company's percentages are correct.

H1: The bus company's percentages are incorrect.

Test statistic:

χ² = ∑ ((O-E)²/E) = 1.6726

df = n-1 = 2

p-value = CHISQ.DIST.RT(1.6726, 2) = 0.4333

Decision:

p-value > 0.05, Do not reject the null hypothesis.