1. The independent variable is the nature of the parents' relationship. One possible operational definition would be to record whether or not the parents are divorced; this would make 'parents' relationship' a qualitative variable. The dependent variable would be attitude towards marriage, which could be measured on a 10-point scale. This would make attitude a quantitative variable. The experiment is observational.

1. a) Low arousal: Mean = 5.1; Median = 5; Mode = 5; s² = 2.32; s = 1.52
   Med Arousal: Mean = 17; Median = 17.5; Mode = 18; s² = 2.0; s = 4.0
   High Arousal: Mean = 7.7; Median = 8; Mode = 8; s² = 1.57; s = 1.25
   
   b)The data suggest that a moderate level of arousal is optimal. This idea has been borne out by many experiments and is often referred to as the Yerkes-Dodson law.



2. These data suggest that all three distributions are symmetrical (or at least pretty darn close).

1. 80% CI = 820 +/- [1.28 *(20/sqrt(100)] = 820 +/- 2.56 = [817.44 - 822.56]
2. 99% CI = 820 +/- [2.58 *(20/sqrt(100)] = 820 +/- 5.16 = [814.84 - 825.16]

1. 95% CI = 12 +/- [2.776 *(1.58/sqrt(5)] = 12 +/- 1.96 = [10.04 - 13.96]
2. 99% CI = 12 +/- [4.604 *(1.58/sqrt(5)] = 12 +/- 1.96 = [8.75 - 15.25]
3. The Confidence Interval gets wider as the level of confidence increases.

1. Ho: Mu = 1250 Ha: Mu does not = 1250
   RR: t < -2.00; t > 2.66
   Zobs = x-MUo / [s/(sqrt(n)] = 1284 - 1250 / [90/sqrt(61)] = 2.95
   tobs falls in the rejection region. Therefore, we would REJECT the  null. Interpretation: Athletes are drawn from a population with a higher mean average SAT score than the general student body.
2. Ho: Mu = 1 Ha: Mu does not = 1
   RR: t < 2.776; t > 2.776
   tobs = x-MUo / [s/(sqrt(n)] = 1.34 -1 / [.5 /sqrt(5)] = 1.52
    tobs does not fall in the rejection region. Therefore, we would FAIL  TO REJECT the null. Interpretation: I should probably stop taking the drug because there is no evidence that nandrolene affects my publication  rate (and there might be some negative side effects).
3. Ho: Mu = 2.6 Ha: Mu does not = 2.6
   RR: t < -2.445; z > 2.445
   M = 4
   s = 2.08
   tobs = M-MUo / [s/(sqrt(n)] = 4 - 2.6 / [2.08/sqrt(7)] = 1.78
   tobs does not fall in the rejection region. Therefore, we would FAIL TO REJECT the  null hypothesis. Interpretation: I do not remember more or fewer dreams than the average person according to Susan Boyle's book.

1. Ho: Uhp = Ue Ha: Uhp is not equal to Ue tcrit = 1.98 (use value for df = 120)
   SE = sqrt[(50²/100) + (75²/100)] = 9.01
   t = 165-150 / 9.01 = 1.66
   tobs is less than tcrit. Therefore, we would fail to reject the null.
   Interpretation: neither printer is faster than the other.
   1a) tcrit = 1.658. 1.66 is greater than 1.658 so we would reject the null.
2. 95% CI = 15 +/- 1.98 * 9.01 = [-2.84-32.84]. Yes, it makes sense. We failed to reject the null, which is consistent with the fact that the interval includes the value 0.

1. tobs = (69.4 - 66.2) / (12.2 /sqrt(120))

   = 2.87 < tcrit (2.66)

   Therefore, you would reject the null and conclude that the jingle improved the consumers' moods.

2. 99% CI = 3.2 +/- 2.66 * (12.2 /sqrt(120))

   = 3.2 +/- 2.66 * 1.11

   = 3.2 +/- 2.95

   = 3.2 +/- [.25 - 6.15]

   In other words, we would expect that the average person's mood would increase by between approximately .25 and 6.15 points after listening to the jingle.

1. 	East	West	Same
   	2	6	1
   	1	4	0
   	3	6	1
   	3	8	1
   	2	5	0
   	4	7	0
   				Sum
   Sum (x)	15	36	3	54
   Sum (x2)	43	226	3	272
   (T2)/n	37.5	216	1.5	255
   Sum (x2)-(T2)/n	5.5	10	1.5	17

   
   

   Source	SS	df	MS	F
   Between	93	2	46.5	41.03
   Within	17	15	1.13
   Total	110	17

   
   The observed value of F exceeds the critical value. Therefore, we would reject the null and conclude that direction of travel does influence the severity of jet lag.


2. Tukey's HSD = q * sqrt(MSE/n)
     = 3.67 * sqrt (1.13/6)
     = 3.67 * .434     =      1.59
   Based on Tukey's HSD, I would conclude that traveling West is more difficult than either traveling East or staying within the same time zone. Traveling East is also more difficult than staying within the same time zone.

1. In a repeated measures experiment, the same units (people) serve as observations (subjects) for each level of the independent variable. In a between subjects experiment, each level of the independent variable consists of a unique set of observations.
2. The difference between RM- and BS-ANOVA relates to the error term. In a BS-ANOVA, the error term consists of all of the variability within each level of the independent variable. In a RM-ANOVA, the variability within each level of the IV is derived from two sources: chance variation (error), and differences between the subjects that comprise the sample. Thus, we can reduce the error term in our F-ratio by partitioning out the variability due to individual differences. Reducing the error term in our F-ratio increases the value of F, which increases the chance that we will reject the null hypothesis. Because reducing the error term increases the probability of rejecting the null without affecting alpha, we say that our test is more powerful.

1. a) See above.
   b) 3 (b/c dfA= 2)
   c) 3 (b/c dfB = 2)
   d) 108 (b/c dfTotal = 108)
   e) 12 (N/treatments = 108 / 9)
   f) A is significant; B is not significant; the interaction is significant.
2. 
   

   	Old AM	Old PM	Young AM	Young PM
   Mean	4	2	4	9
   Sum (x)	24	12	24	54	114
   Sum(x2)	110	28	118	498	754
   SS	14	4	22	12	52
   (G2)/N					541.5
   (T2)/n	96	24	96	486	702
   Sum(x2)-(T2)/n	14	4	22	12	52

   
   
   

   Source	SS	df	MS	F	Fcrit
   Model	160.5	3	53.5	20.58	3.10
   Error	52	20	2.6
   Total	212.5	23

   Because the observed value of F exceed the critical value, we conclude that at least one of our treatment means differs from the others (F 3, 20) = 20.58, MSE = 2.6). Now, on to test the main and interaction effects.
   
   

   	Older	Younger	AM	PM
   Age	108	507			615
   Test Time			192	363	555

   
   
   

   Source	SS	df	MS	F	Fcrit
   Age	73.5	1	73.5	28.27	4.35
   Test Time	13.5	1	13.5	5.19
   Age x Test Time	73.5	1	73.5	28.27

   
   The main effect of Age was significant (F (1, 20) = 28.27, MSE = 2.6). Younger adults performed significantly better than older adults. The main effect of Test Time was also significant (F (1, 20) = 5.19, MSE = 2.6). the subjects performed better in the afternoon than in the morning. Finally, the interaction effect was also significant (F (1, 20) = 28.27, MSE = 2.6). Whereas older adults tended to perform better in the morning, younger adults tended to perform better in the afternoon. Moreover, older adults performed just as well as younger adults in the morning, but performed much worse than younger adults when tested in the afternoon.
   

1. 1. Calculate the regression equation.
      SP = S(xi)(yi) - [S(xi)S(yi)] / n
      

      = 15,013 - [(109.5*11,087)/77]

      = 15,013 - 15766.58 = -753.58

      
      

      SSx = Sxi² - [(Sxi)² / n]

      = 246.25 - [(109.52)/77]

      = 246.25 - 155.72 = 90.53
      

      B1 = SP / SSx

      = -753.58 / 90.53 = -8.32

      B0 = My - B1 (Mx)

      = (11,087/77) - (-8.32)*(109.5/77)

      = 143.98 - (-8.32)*1.42 = 155.82

      y = 155.82 - 8.32*x
   
   
   2. The slope tells us the change in y for a unit change in x. Therefore, if you ate one additional serving of vegetables, I would expect you to lose 8.32 pounds.
      
   3. Use the regression equation with 3 substituted for x.

      y = 155.82 - 8.32*x

      y =155.82 - 8.32*(3)

      y =155.82 - 24.96

      y =130.86

   4. Performing the hypothesis test.

      SSy = Syi2 - [(Syi)2 / n]

      = 1666795 - (110872/77)

      = 1666795 - (122921569/77)

      = 1666795- 1596384.012987

      = 70410.987013

      SSE = SSy - b1(SP)

      = 70410.987013 - (-8.32 *-753.58)

      = 70410.987013 - 6269.7856

      = 64141.201413

      s2 = MSE = SSE / (n-2)

      = 64141.201413 / (77-2) = 855.22 s = sqrt(855.22) = 29.24

      t = b1 - 0 / s / sqrt(SSx)

      = -8.32 / (29.24 / sqrt(90.53))

      = -8.32 / (29.24 / 9.515)

      = -8.32 / 3.073

      = -2.71

      Reject the null. We conclude that there is a significant relationship between weight and vegetable consumption.

   5. Calculating the correlation coefficient

      SSy = Syi² - [(Syi)² / n]

      = 1,666,975 - (11,0872/77)

      = 1,666,795 = 1,596,384.01 = 70,410.99

      r = SP / [sqrt(SSx)(SSy)]

      = -753.58/(sqrt(90.53*70,410.99))

      = -753.58 /sqrt(6374306.65) = -.30

   r² = -.302 = .09.

   Although vegetable consumption does help us predict weight, it can only explain 10% of the variance in our dependent measure. Clearly, other factors are involved

1. y = 6.967 + .596(x)
2. .596 hours later.
3. y = 14.119 (approximately 2:00 AM)
4. The tobs = 5.212; because my p-value is less than .05, I would reject the null and conclude that weekday bedtime is a significant predictor of weekend bedtime.
5. r = .53; therefore, r2 = .281. This suggests that we can explain almost 30% of the variance in weekend bedtimes if we know weekday bedtimes. That is a reasonably high value for "real" data.

Solution

SSx = 350385 - (1667²/8)

= 3023.875

SSy = 555 - (61²/8)

= 89.875

SP = 12308 - (1667*61/8)

= -402.875

r = SP / sqrt(SSx*SSy)

= -402.875 / sqrt(3023.875*89.875)

= -.7728

Thus, there is a strong negative relationship between the speed of response and accuracy. In other words, faster responses tend to produce more errors.

2) Yes. The correlation coefficient is significant; r (6) = -.773, p = .025).

1. Yes, the overall model will help SMS predict success in the fraternity. This conclusion is based on the fact that the omnibus ANOVA leads us to rejct the null (F (6, 42_ = 6.00, MSE = 5.00, p = .0125).
2. R-squared = 180/360 = .50
3. y = 22 - .16(GPA) + 2.17(Math-GPA) + 4.81(Protect) + 6.03(Trek) + .88(T-shirts) - 3.25(Friends)
4. y = 22 - .16(3.20) + 2.17(3.80) + 4.81(2) + 6.03(2) + .88(12) - 3.25(2)
5. y = 22 - .512 + 8.246 + 9.62 + 12.06 + 10.56 - 6.5 y = 55.474
6. Therefore, we would expect your brother to be admitted to the fraternity.