Question

the publisher decided to introduce the book to the market with five different colour, each having a different colour. the first edition includes 300 copies the following frequency of each colour is observed

red 52

blue 77

green 62

pink 64

orange 45

the publisher wants to check if there is evidence that fhe number of each equal is not equal

1) write the hypothesis of the test for this case

2) based on data, what is the test stat for this test using 0.05 significance

3)based on data what is the critical value for this test using 0.05 level signifance

Answer 1

Answer: The publisher decided to introduce the book to the market with five different colour, each having a different colour. the first edition includes 300 copies the following frequency of each colour is observed.

Solution:

Proportion p = 1/5

1) The null and alternative hypothesis test:

Ho: p1 = p2 = p3 = p4 = p5 = 1/5

Ha = atleast one proportion is not equal.

2) From the given data we used chi square test statistic χ2 at α = 0.05 significance level.

n = 300

Expected frequency = n*p

Expected frequency = 300 * 1/5

Expected frequency = 60

Sno	Observed value (O)	Expected value (E)	(O-E)^2/E
Red	52	60	1.0667
Blue	77	60	4.8167
Green	62	60	0.0667
Pink	64	60	0.2667
Orange	45	60	3.7500
Total	300	300	9.9668

Test statistic χ2 = Σ(O-E)^2/E

Test statistic χ2 = 9.9668

3) Critical value:

Degree of freedom, df = k-1

df = 5-1 = 4

Critical value of χ2 at α = 0.05 and df = 4

Critical value of χ2 = 9.488

Since, test statistic χ2 (9.9668) > critical value of χ2 (9.488)

Conclusion:

Reject the null hypothesis Ho. There is sufficient evidence to conclude that the number of each is not equal.