Question

The table below lists the number of games played in a yearly​ best-of-seven baseball championship​ series, along with the expected proportions for the number of games played with teams of equal abilities. Use a 0.05 significance level to test the claim that the actual numbers of games fit the distribution indicated by the expected proportions.

Games Played	4	5	6	7
Actual contests	1818	2222	2222	3939
Expected proportion	two sixteenths216	four sixteenths416	five sixteenths516	five sixteenths516

Determine the null and alternative hypotheses.

Upper H 0H0​:

▼

Upper H 1H1​:

▼

Calculate the test​ statistic,

chi squaredχ2.

chi squaredχ2equals=nothing

​(Round to three decimal places as​ needed.)

Calculate the​ P-value.

​P-valueequals=nothing

​(Round to four decimal places as​ needed.)

What is the conclusion for this hypothesis​ test?

A.

RejectReject

Upper H 0H0.

There is

sufficientsufficient

evidence to warrant rejection of the claim that the actual numbers of games fit the distribution indicated by the expected proportions.

B.

Fail to rejectFail to reject

Upper H 0H0.

There is

insufficientinsufficient

evidence to warrant rejection of the claim that the actual numbers of games fit the distribution indicated by the expected proportions.

C.

Fail to rejectFail to reject

Upper H 0H0.

There is

sufficientsufficient

evidence to warrant rejection of the claim that the actual numbers of games fit the distribution indicated by the expected proportions..

D.

RejectReject

Upper H 0H0.

There is

insufficientinsufficient

evidence to warrant rejection of the claim that the actual numbers of games fit the distribution indicated by the expected proportions.

Answer 1

Hypotheses are:

H0: The number of games played in a yearly​ best-of-seven baseball championship​ series are as per the expected proportions.

Ha: The number of games played in a yearly​ best-of-seven baseball championship​ series are not as per the expected proportions.

Following table shows the calculations for test statistics:

	O	p	E=p*101	(O-E)^2/E
	18	0.125	12.625	2.288366337
	22	0.25	25.25	0.418316832
	22	0.3125	31.5625	2.897153465
	39	0.3125	31.5625	1.75259901
Total	101	1	101	7.356435644

The test statistics is:

Degree of freedom:

df =4-1 = 3

The p-value using excel function "=CHIDIST(7.356,3)" is 0.0614

Since p-value is greater than 0.05 so we fail to reject the null hypothesis.

Correct option:

B. Fail to reject H0.

There is insufficient evidence to warrant rejection of the claim that the actual numbers of games fit the distribution indicated by the expected proportions.