STAT 201 Assignment 6

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1 STAT 201 Assignment 6 Partial Solutions 12.1 Research question: Do parents in the school district support the new education program? Parameter: p = proportion of all parents in the school district who support the new education program Sample estimate: p = 104/ (a) Research question: How many messages do subscribers of this internet provider receive on any given day? Parameter: µ = mean number of s per subscriber of this internet provider received on the day before the survey Sample estimate: µ = Research question: Is herbal tea or throat lozenge more effective in relieving sore throats? Parameter: p h p l = difference in proportion of relieved throats between the population of all sore throat sufferers who d drink herbal tea and that of all sore throat sufferers who d take throat lozenges Sample estimate: p h p l = difference in proportion between sample from tea drinkers and sample from lozenge poppers 12.4 Research question: Do students from the university spend more time studying or socializing? Parameter: µ d = mean of paired differences of # hours per week spent studying and # hours per week spent socializing for all students at the university Sample estimate: D = mean of paired differences of # hours per week spent studying and # hours per week spent socializing for the 100 selected students 12.5 paired two variables measured on each person sampled from the same population 12.6(a) paired (population of couples) (b) independent (population of teachers, population of plumbers) (c) independent (population of HS grads, population of college grads) 12.9(a) SE(X) = s x / n x = 2/ 64 (b) n = 64 quite large: use CLT and pretend s very close to σ. x ± 2 s x / n x = 27.5 ± 2 2/ 64 = (27.0, 28.0) We are about 95% confident that the mean foot length for all men is somewhere between 27.0 and 28.0 cm (a) SE(Y ) = s y / n y = 2/ 100 smaller because sample is larger

2 (b),(c) Both samples are large: use CLT (i.e. X Y approx. normal), and pretend s σ for both groups. Samples are obviously independent: X Y approx N(µ x µ y, SE(X Y ) = 4/64 + 4/100 σ 2 x /n x + σ 2 y /n y) N(µ x µ y, Approx. 95% C.I. for difference between means: ± 2 4/64 + 4/100 = (2.86, 4.14) s 2 x /n x + s 2 y /n y) We are about 95% confident that on average, men s and women s foot sizes differ by somewhere between 2.86 and 4.14cm paired observations (before-after); n = 50 large: D approx N(µ d, σ d / 50) N(µ d, s d / 50). So compute d = after-minus-before rates, s d = sample SD for the differences, SE(D) = s/ (a) SE(X) = s/ n = 3.7/ 42 This is the typical estimation error when we use the sample mean to estimate the mean weight loss if all men went on the diet plan. (b) n = 42 large: X approx N(µ, σ/ n) N(µ, s/ n) x ± 2 s/ n = 7.2 ± 2 3.7/ 42 = (6.06, 8.34) We are about 95% confident that for all men who go on this diet, they d lose, on average, somewhere between 6.06 and 8.34 kg Her population is the class itself no sampling involved she knows all population parameter values (a) x y = statistic: came from sample (b) SE(X) = s x / n x = 1.51/ 116 SE(Y ) = s y / n y = 1.87/ 59 (c) We are about 95% confident that, on average, the difference between females and males daily TV hours is somewhere between and 0.97, which includes 0 (i.e. no difference). So we can t be so sure gender makes a difference in daily TV hours. (d) Both samples (independent) are large: X Y approx N(µ x µ y, σ 2 x /n x + σ 2 y /n y) N(µ x µ y, s 2 x /n x + s 2 y /n y) So C.I. formula used 2 as multiplier and SE = / / Both samples (independent) are large: approx p 1 p 2 N(p 1 p 2, p 1 (1 p 1 )/n 1 + p 2 (1 p 2 )/n 2 ) N(p 1 p 2, p 1 (1 p 1 )/n 1 + p 2 (1 p 2 )/n 2 ) Approx. 95% C.I. is p 1 p 2 ±2 p 1 (1 p 1 )/n 1 + p 2 (1 p 2 )/n 2 = ± / /300 = (0.26, 0.42) We are about 95% confident that the difference in proportion of YES s between all men and all women is somewhere between 26% and 42% (i.e. quite different).

3 12.16 p 1 approx N(p 1, p 1 (1 p 1 )/n 1 ) N(p 1, p 1 (1 p 1 )/n 1 ) Approx. 95% C.I.: 0.63 ± /300 = (0.57, 0.69) We are about 95% confident that between 57% and 69% of all men would say YES to the question All n s are small can t use CLT or pretend s σ. So need to assume population histograms are reasonably bell-shaped, so that we can use the t multipliers for our C.I. s: x ± t s/ n where t is from t-distribution with n 1 degrees of freedom. (a) 76 ± / 9 (b) 76 ± / 9 (c) 76 ± / 16 (d) 76 ± / 16 (e) 100 ± / confidence level (longer C.I. for higher level), n (shorter C.I. smaller estimation error for larger n), σ (longer C.I. larger estimation error for larger σ) n x = 25, n y = 23, x = 4.5, s x = 1.6, y = 8.1, s y = 1.8 Samples not too large: assume both populations are normal and use t-distribution for C.I. s. (a) 4.5 ± / 25 = (3.8, 5.2) (b) 8.1 ± / 23 = (7.3, 8.9) (c) The C.I. s don t overlap: we are quite confident the two groups differ in symptom duration. (d) *********IGNORE for now********** ± / 23 = (7.04, 9.16) We are 99% (i.e. extremely) confident that cold sufferers would take, on average, somewhere between 7.04 and 9.16 days (more than a week) to recover if they all took placebo lozenges (a) present: we use random sample to infer about the mean amount spent by all students at the university on the semester s books (b) s/ n = 96.1/ 12 (c) 11 degrees of freedom; t = C.I. s half-width SE = ( )/ = (d) We are about 95% confident that on average, the university students spent somewhere between $224 and $346 on this semester s books. (e) t = 1.80 Approx. 90% C.I.: ±

4 12.22 Estimate population mean and SD by sample mean and SD. So if population s mean expense were $284.9 with an SD of $96.1, and that the histogram is relatively bell-shaped, then ± = (92.7, 477.1) would cover about 95% of the university s students (a) observed values are those of random sample from population of all drivers who d pass by the sign (b) n = 16, x = , s = n not large: assume population histogram is reasonable bell-shaped and use t-distribution. Approx. 95% C.I. for population mean: ± / 16 = (430.25, ). We are about 95% confident that on average, drivers who pass by the sign can see it from a distance of somewhere between and feet (a) 1.8 ± / 20 = (0.82, 2.78) We are about 95% confident that on average, 18- to 29-year-old women wish to be somewhere between 0.82 and 2.78 inches taller than they actually are. (b) Population of the wished differences is approximately normal. (Alternatively: 18- to 29-year-old women s heights are approximately normal; their wished heights are approximately normal (so that the wished differences are automatically approximately normal).) (c) The question posed to the women in the sample already assumed they wanted to be taller (a) We are quite confident the two populations are different. (b) We can t say with much confidence that the two populations are different n x = 25, n y = 23, x = 4.5, s x = 1.6, y = 8.1, s y = 1.8 (a) n s not too large, need t-distribution; so assume both populations are relatively normal. x y = = 3.6 Unpooled SE(X Y ) = s 2 x /n x + s 2 y /n y = / /23 (b) t = 2.07 for df = So approx. 95% C.I. is 3.6 ± / /23 = (2.58, 4.62). (c) YES: the 95% C.I. doesn t contain Assume both populations are relatively normal with equal σ (which is supported by the closeness of the sample SD s). Pooled estimate of σ: (n x 1)s 2 x s p = + (n y 1)s 2 y 24(1.6)2 + 22(1.8) = 2 = 1.70 n x + n y SE(X Y ) = s p 1/nx + 1/n y = /25 + 1/23 = 0.49 t for df = n x + n y 2 = 46. Approx. 95% C.I. is 3.6 ±

5 12.28(a) estimate: x y = = 0.90 estimation error: SE(X Y ) = s 2 x /n x + s 2 y /n y = / /10 = 1.18 Small samples: assume both populations relatively normal and use t-distribution. t = 2.07 for df=22 Approx. 95% C.I. for difference in mean between populations is 0.90 ± (b) We are about 95% confident that on average, the saliva testosterone levels between MD s and professors differ by somewhere between and 3.34 ng/dl. (c) We can t say with much confidence that they re different Let the NO s be X s and YES s be Y s. x = 72, s x = 9.13, n x = 8, y = 65.33, s y = 7.49, n y = 12 Samples small: assume both populations are relatively normal; not wise to use unpooled SE, so also assume populations have common σ (indicated by relative closeness of sample SD s). Pooled estimate for σ: t = 2.10 for df = 7+11 s p = 7(9.13)2 + 11(7.49) = Approx. 95% C.I. for difference between population mean pulse rates: ± = ( 10.5, 23.8) 12.30(a) Let p 1 be the proportion for the younger population; p 2, the older. p 1 = 653/1600, p 2 = 995/2122, p 1 p 2 = (b) Samples (independent) are huge, so CLT will work well and p s very close to true p s: p 1 p 2 approx N(p 1 p 2, Approx. 95% C.I. for p 1 p 2 : (653/1600) (947/1600) ± p 1 (1 p 1 ) + p 2(1 p 2 ) p 1 (1 p 1 ) ) N(p 1 p 2, + p 2(1 p 2 ) ) n 1 n 2 n 1 n 2 (995/2122) (1127/2122) 2122 = ( 0.094, 0.028) We are about 95% confident that the younger and older women populations differ in proprotion who had experienced episodic headaches by somewhere between -9.4% and 2.8% (i.e. we can t say with confidence that they are different) (a) We are about 95% confident that the difference in proportion of those who had experienced episodic headaches between the more educated group and less educated group is somewhere between 6.8% and 10.4% (i.e. quite confident there is a difference). (b) (See rationale for Question 12.30(b).) ± / / The two groups in the sample didn t come from two separate populations. (In fact, it is just one sample split into 2 according to their value for the response variable, i.e. preference.)

6 12.33 p f = 0.611, p m = (a),(c) Approx. 95% C.I.: ± / /61 = (0.032, 0.034) We are about 95% confident that the female and male proportions of YES s is between 3.2% and 3.4% (i.e. quite confident there is difference). (b) college statistics students 12.34(a) Use z (or t with infinite degrees of freedom) for 99% confidence: ± / /61 (b) longer (c) narrower (lower confidence bigger chance for the C.I. to miss the truth) n large: X approx N(µ, σ/ n) N(µ, s/ n) Approx. 95% C.I. is 0.92 ± / 235 = (0.85, 0.99). We are about 95% confident that the mean composite score for all PMS sufferers who would take the placebo is between 0.85 and (a) Samples are huge and obviously independent: X Y approx N(µ x µ y, Approx. 95% C.I. is σ 2 x /n x + σ 2 y /n y) N(µ x µ y, ± / /231 = ( 0.079, 0.119) s 2 x /n x + s 2 y /n y) We are about 95% confident that for the baseline group 7 days prior to treatment, all PMS sufferers under placebo and under treatment differ in their mean composite score by somewhere between and (i.e. can t say with confidence there is difference). (b) Can t expect drug to take effect 7 days before treatment actually began! relative risk of equal to 1 is absolutely same risk (a) quite confident the risks are different (b) can t say with confidence there is difference Assuming the C.I. s are of high confidence level, we see neither contains 1, i.e. in both cases we are quite confident the risks are different.

Section I Questions with parts

Name: SFU ID: Stat 201 Midterm Solutions page 1 Section I Questions with parts Most parts in these questions are related. For a part that you can t do, write Can t do -- suppose answer is and fill in some