Question

A) In this problem, assume that the distribution of differences is approximately normal. Note: For degrees of freedom d.f. not in the Student's t table, use the closest d.f. that is smaller. In some situations, this choice of d.f. may increase the P-value by a small amount and therefore produce a slightly more "conservative" answer.

At five weather stations on Trail Ridge Road in Rocky Mountain National Park, the peak wind gusts (in miles per hour) for January and April are recorded below.

Weather Station	1	2	3	4	5
January	137	120	128	64	78
April	108	113	102	88	61

What is the value of the sample test statistic? (Round your answer to three decimal places.)

B)

In this problem, assume that the distribution of differences is approximately normal. Note: For degrees of freedom d.f. not in the Student's t table, use the closest d.f. that is smaller. In some situations, this choice of d.f. may increase the P-value by a small amount and therefore produce a slightly more "conservative" answer.

Do professional golfers play better in their last round? Let row B represent the score in the fourth (and final) round, and let row A represent the score in the first round of a professional golf tournament. A random sample of finalists in the British Open gave the following data for their first and last rounds in the tournament.

B: Last	70	66	71	71	71	72	68	68	74
A: First	68	69	61	71	65	71	71	71	71

What is the value of the sample test statistic? (Round your answer to three decimal places.)

Answer 1

A)

S. No	January	April	diff:(d)=x1-x2	d2
1	137	108	29	841.00
2	120	113	7	49.00
3	128	102	26	676.00
4	64	88	-24	576.00
5	78	61	17	289.00
	total	=	Σd=55	Σd2=2431
	mean dbar=	d̅     =	11.000
	degree of freedom =n-1                            =		4
	Std deviaiton SD=√(Σd2-(Σd)2/n)/(n-1) =		21.365861
	std error=Se=SD/√n=		9.5551
	test statistic            =	(d̅-μd)/Se         =	1.151

B)

S. No	B	A	diff:(d)=x1-x2	d2
1	70	69	1	1.00
2	66	69	-3	9.00
3	71	61	10	100.00
4	71	71	0	0.00
5	71	65	6	36.00
6	72	71	1	1.00
7	68	71	-3	9.00
8	68	71	-3	9.00
9	74	71	3	9.00
	total	=	Σd=12	Σd2=174
	mean dbar=	d̅     =	1.333
	degree of freedom =n-1                            =		8
	Std deviaiton SD=√(Σd2-(Σd)2/n)/(n-1) =		4.444097
	std error=Se=SD/√n=		1.4814
	test statistic            =	(d̅-μd)/Se         =	0.900