Question

At a certain hotel, guests have come from the following locations: Use Table 3.

Category	North America	Asia	Other	TOTAL
Proportion	60	30	10	100

Category	North America	Asia	Other
Frequency	238	107	45

a.	Choose the appropriate alternative hypothesis at H0: p1 = 0.60, p2 = 0.30, and p3 = 0.10.
	All population proportions differ from their hypothesized values. At least one of the population proportions differs from its hypothesized value.

b.	Calculate the value of the test statistic at H0: p1 = 0.60, p2 = 0.30, and p3 = 0.10. (Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.)

Test statistic

c.	Approximate the p-value.
	p-value < 0.005 p-value > 0.100 0.010 < p-value < 0.025 0.025 < p-value < 0.050 0.050 < p-value < 0.100

d.	At 5% significance level, can we reject H0: p1 = 0.60, p2 = 0.30, and p3 = 0.10?
	Yes since the p-value is less than α. Yes since the p-value is more than α. No since the p-value is less than α. No since the p-value is more than α.

NEED DONE IN 30 TIMED HW

Answer 1

a)

Null hypothesis H0 : p₁ = 0.60, p₂ = 0.30, and p₃ = 0.10

Alternative hypothesis H1 : At least one of the population proportions differs from its hypothesized value.

b)

Test statistic :

Where, Oi is observed frequency.

Ei is Expected frequency.

Ei = N*Pi

Calculation :

Test Statistic = 1.85  

c ) P-value :

Using Excel function ,   =CHIDIST( x , df )    

df = 3 - 1 = 2 , x = test statistic

P-value = CHIDIST( 1.85 , 2 ) = 0.3965

i.e p-value > 0.100

d )

Decision about null hypothesis -

Rule : Reject null hypothesis if p-value less than significance level

Significance level = = 5% = 0.05

It is observed that p-value is greater than significance level = 0.05

fail to reject null hypothesis.

Correct option : No since the p-value is more than α.