Question

This Question: 1 pt 1 of 14 (1 complete) This Test: 14 ptsp Question Help A book claims that more hockey players are bom in January through March than in October through December. The following data show the number of players selected in a draft of new players hockey league according to their birth month. Is there evidence to suggest that hockey players' birthdates are not uniformly distributed throughout the year? Use the level of significance -0.01. 3 Click the loon to view the table War TU TYPOLIT VE TYPES i Birth months of players O A. Ho: The distribution of hockey players' birth months is uniformly distributed H,: The distribution of hockey players' birth months is not uniformly distributed O B. Me: The distribution of hockey players' birth months is uniformly distributed. H,: More hockey players are born in the first half of the year than the second half O C. He: The distribution of hockey players' birth months is uniformly distributed H,: More hockey players are born in January March than October December OD. Ho: The distribution of hockey players' birth months is not uniformly distributed Hy! The distribution of hockey players' birth months is uniformly distributed Birth Month Frequency Hockey League Players' Birth Month January March April-June July-September October-December 160 62 33 Print Done Compute the expected counts for each birth month. The total number of hockey players is 179 Birth Month January March April-June July-September October-December Observed Count Expected Count 60 62 33 24 (Round to two decimal places as needed.) What is the test statistic? x - l (Round to three decimal places as needed.) What is the P-value of the test? P-value = (Round to three decimal places as needed.) Click to select your answer Il auto.proctoru.com is sharing your screen. Stop sharing

Answer 1

H₀: The distribution of hockey players’ birth months is uniformly distributed.

H₁: The distribution of hockey players’ birth months is not uniformly distributed.

The output is:

observed	expected
60	44.750
62	44.750
33	44.750
24	44.750
179	179.000

Frequency	Expected
60	=0.25*179
62	=0.25*179
33	=0.25*179
24	=0.25*179
=SUM(60+62+33+24)	=SUM(44.75+44.75+44.75+44.75)

The test output is:

observed	expected	O - E	(O - E)² / E	% of chisq
60	44.750	15.250	5.197	21.17
62	44.750	17.250	6.649	27.08
33	44.750	-11.750	3.085	12.57
24	44.750	-20.750	9.622	39.19
179	179.000	0.000	24.553	100.00
24.553	chi-square
3	df
1.91E-05	p-value

The test statistic is 24.553.

The P-value is 0.000.

No, because the calculated p-value is less than the given alpha level of significance.

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