Key to CFU on Chapters 13 and 14

Statistics / Mr. Hansen
4/19/2004

Check one:

Name: ____________KEY______________
c Regular time (36 minutes)
c Extended time (54 minutes)

CFU on Chapters 13 and 14

Instructions and Scoring.

Show all work and answers on separate sheets.
Show as much work as time provides. Justify all steps except for algebraic simplification or area calculations. For full credit, you should show the details of s.e. and m.o.e. calculations to the extent possible.
You will not earn full credit for a correct answer without adequate justification, or a correct answer without correct notation.

1.	A women’s group has contracted with an independent research organization to gather evidence to support the notion that all men are slobs. To this end, they have analyzed the following data regarding education level and likelihood that men wash their hands after using the bathroom:

	Education level	% of hand washers
	below H.S. H.S. diploma some college college graduate some graduate sch. postgraduate degree	44% 38% 36% 58% 48% 50%

(a)	Formulate a test for independence between education level and percentage of hand washing. Assume that the men used were a random sample and that there were 50 in each group. Raise your hand before you start punching buttons on your calculator.
(b)	Show the work for the calculation for the expected entry in the fourth row and first column.
(c)	Carry out the test, showing all steps and writing a conclusion.

(a)	2-way table of counts:

	Educ.		Wash		No Wash
	< H.S.		22		28
	H.S.		19		31
	some coll.		18		32
	B.S.		29		21
	some grad. sch.		24		26
	master’s +		25		25

	H₀: Educ. & hand-washing behav. are independent
	**H_a: There is some assoc. betw. educ. & hand-washing behav.**

(b)	E₄₁ = (row 4 tot.)(col. 1 tot.)/(grand tot.) = 50 · 137 / 300 = 22.833
	Expected matrix (by calc.):
				22.833	27.167
				22.833	27.167
				22.833	27.167
				22.833	27.167
				22.833	27.167
				22.833	27.167

(c)	Assumptions: ü SRS (random was given, must assume SRS) ü all expected counts ³ 1 (see above) ü no more than 20% of expected counts < 5 (see above)

	Test statistic: c² = S (obs. – exp)²/exp. = (22 – 22.833)²/22.833 + (28 – 27.167)²/27.167 + . . . = 6.677 p = .246 Conclusion: There is no evidence (c² = 6.677, df = 5, p = .246) of an association between education and type of male hand-washing behavior. The variations seen in the 2-way table of observed counts could plausibly be explained by chance alone.

2.	Researchers are interested in the relationship between time remaining to graduation and seniors’ happiness level, on a scale from 1 to 10. On a random sample of school days, the following data were gathered:


	Time Left (Days)	Form VI Happiness Level
	90 80 70 60 45 40 30 20 10 5	1 3.4 5 4.8 5.9 6.5 7.2 7 9.92 9.998

(a)	Describe the relationship between the variables, using words that indicate the context of the problem. Provide at least one diagram and at least one piece of quantitative evidence to support your claim.
(b)	Using time remaining as a predictor of happiness level, compute the slope of the line of best fit.
(c)	Interpret your answer to part (b) in the context of the problem, using words such as “time remaining” and “happiness level.” Your answer should make sense to someone who has studied little or no statistics.
(d)	Compute a 90% confidence interval for the slope requested in part (b).
(e)	Perform a significance test for the proposition that the true linear correlation coefficient is nonzero. Show all steps.

(a)	There is a strong negative linear relationship (r = –0.962) between days remaining to graduation and Form VI happiness as measured on a 10-point scale. A residual plot [should be shown here] reveals no obvious pattern to doubt the validity of the linear fit.

(b)	slope = b₁ = –.08995 [No work required. However, if time permits, you could justify by saying yhat = 10.1196 – .08995x by calc., where x = days to graduation and yhat = predicted happiness.]

(c)	For each additional day remaining to graduation, our model predicts a drop in happiness level of .08995 units on a 10-point scale. [Also acceptable: For each 1-day decrease in time remaining to graduation, our model predicts a happiness increase of .08995 units.] [Note: The word “model” or “predicts” must be present in the wording for full credit. Simply saying that there is a .08995 happiness increase for each day closer to graduation is not correct.]

(d)	*s_b₁ = b₁ / t = –.08995/(–9.96. . .) = .0090291844 C.I. = est. ± (crit. val.)(s.e.) = –.08995* ± *(t)(s_b₁) = –.08995** ± 1.86(.0090291844) = (–.1067, –.0732) We are 90% confident that the true LSRL slope is between –.1067 and –.0732.

(e)	Let r = true lin. correl. coeff. H₀: r = 0 H_a: r ¹ 0 Assumptions: ü random sample of data points ü true assoc. appears to be linear since resid. plot shows no pattern [must show resid. plot] ü resids. do not vary w/ change in x [must show resid. plot] ü resids. normally distributed [must show stemplot, histogram, or NQP] Test statistic: t = –9.96228 by calc. p = .00000873 by calc. (2-sided) Conclusion: There is strong evidence (t = –9.962, df = 8, p < .0001) that the true value of the lin. correl. coeff. is nonzero.