TMTH 3360

APPLIED TECHNICAL STATISTICS

Selected Answers to Homework Problems

Note that these solutions are from the Instructor's Manuel and have not been checked. They should probably be correct, but if you have a problem, be sure to contact me.

Chapters

1	2	3	4	5
6	7	8	9	10

Problem #
Chapter 1		a	b	c	d
1.2		quantitative		qualitative
1.4			age	quantitative
1.9		see text		c and d
1.12		political view is qualitative		% of people in each of the 4 categories
1.14		mound shape	1.6 (16)
1.20		Note that the measurements decrease over time.
1.22			Skewed right with a few large measurements
1.33		possibly skewed		not likely skewed
1.36		continuous for positive values	continuous	discrete	discrete
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Chapter 2		a	b	c	d
2.2			4
2.4
2.8
2.14					5/1.55 = 3.23
2.16		68% in [33,39] etc.	75% in [30,42] etc.
2.18		approx. 90%		at most 25% are less than 65
2.20			0.166		% in c) not consistent with the Empirical Rule; caused by gap in the center of the distribution
2.31		s² = 7.873 so s =?
2.34		IQR = 26-22 = 4; suspected outlier = 12
2.38			1.94 & 2.12
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Chapter 3		a	b	c	d
3.2				side-by-side to compare and evaluate difference
3.5
3.12		y = -206.06 + 0.196x
3.19		y = 0.0187 + 0.0042x

Chapter 4		a	b	c	d
4.4		0.21	0.91
4.5			3/4		3/4
4.17		80
4.22		(2019)/(21) = 190
4.26		4124 = 192
4.33		5.720645*10^12
3.32		P(A)*P(B/A) = ?
4.52		0.0004		(0.02)*(0.02)
		Notation P: test is positive for drugs N: test is negative for drugs D: employee is a drug user Dc: the compliment of D (not a user) Information stated in the problem P(P given D) = P(P/D) = .98 P(N given Dc) = P(N/Dc) = .98 Part a P(P given Dc) = P(P/Dc) = 1-.98 = .02 P(both tests fail) = (.02)(.02) [since we assume tests are independent] Part b P(detection given D) = (.98)(.02)+(.02)(.98)+(.98)(.98) = .9996 Part c P(pass both given D) = (.02)*(.02)
4.54		0.4	0.37	0.10	.4+.37-.10=0.67
4.76			0.27		continuous
4.78		0.05	1.85 & .19		0.95
4.93		$1500
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Problem #		a	b	c	d
Chapter 5
5.74		0.32768		0.00672
5.29		0.135335	0.27067		0.036089
5.33			0.1755
5.35		0.271		0.406
5.45			0.1786		0.2857
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Chapter 6		a	b	c	d
6.2		0.8384	0.9974
6.77		-1.96	0.36
6.9		1.28			2.33
6.13		58.3
6.15		8 & 2
6.17		0.1949			3.92; unusual; maybe not a random sample
6.82		.0475	.002266		38.84
6.83		10.2%
6.84		0.0080
6.29		63,500
6.85		0.1251
6.70		1.461; not an unlikely occurrence; no evidence of difference in caloric intake between the groups
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Chapter 7		a	b	c	d
7.24		normal by CLT with mean m and s = 2/sqr(10)	z = -15.8 & 0.0571	21.948
7.28		approximately normal; 95%
7.30		normal approximation	0.0681		0.8638
7.44		7.12 to 7.36	0.7924
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Chapter 8		a	b	c	d
8.4		As the population variance increases, the margin of error also increases
8.6		As the sample size increases, the margin of error decreases.
8.10		+/- 0.098
8.12		.447
8.16		.045
8.20		[12.496,13.704]	2.73+/-0.079
8.23

8.38		[-1.032,-0.368]
8.55		2+2.00
8.69		98
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Problem #
Chapter 9		a	b	c	d
9.2		0.1251		0.0351
9.4		Ha: m>2.3		0.049	calculated z = 2.04; Reject Ho; evidence indicates that the population mean is > 2.3.
9.8		Ho: m=60 Ha: m<60	one tailed	calculate. z = -1.992 cv =-1.645 Reject; Flights unprofitable

9.18		Ho: m₁ - m₂ = 0 Ha: m₁ - m₂ > 0 [may test m₁ - m₂ \| 0] cal z =5.33 and p < 0.001 Reject Ho		[73 - 63] + 2.33 [sqr(3.5225)]
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Chapter 10		a	b	c	d
10.2		t_0.005 =3.055	1.746	2.060	2.998
10.11		3.782+/-0.130
10.19				5.2889	p<0.01
10.38		Measurements are on the same person and thus, not independent. Use a paired test.
10.60		Assume equal variances.		F=1.64; No evidence against assumption.
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