Answers to even HW Stat 217.
Chapter 7.1
2. t* is 2.262 with df = 9, s = 82.792, xbar = 531. (471.77, 590.23)
6. actually matched pairs but the subtraction has been done. a.) Ho: u = 0, Ha: u not equal to 0. Two sided since question is “are they different”. b.) t = (4.8 –0)/(15/sqrt50) = 2.263 For df = 49 gives P that is between 0.01 and 0.02 but multiply that by 2. P<a for sig level of 0.05 so reject, or P>a for sig level of 0.01 so fail to reject. c.) Definitely not certain all stores are up, only on average so some are up, some may be down.
8. No, neg, – 12.4 is an outlier which messes up t calcs more than skewed.
26. t* = 2.080, SE = 0.176 so 95% CI is (1.714, 2.446)
32. a.) df = 4, t* = 2.776, s = 3.94, n = 5 so 95% CI is (50.11, 59.89). b.) you must go back and recalculate all the after data as a % of 98. Get new xbar = 26.33 and s = 7.68 so new 95% CI is (16.79, 35.87).
Chapter 7.2
72. DF (from smaller n) = 44 (use 40 in table) so t* = 2.021. 3.61 – 2.95 +/- 2.021 x 0.0366 which gives (0.586, 0.734)
74. must be pooled! a.) Ho: mu 1 = mu 2, Ha: mu 2 < mu 1. SE = 0.544, t = (13.3 – 12.4)/0.544 = 1.65. Use pooled DF = 40 so p value is between (0.05, 0.10). At alpha = 0.05 we should fail to reject – insufficient evidence to believe there is a difference.
b.) for 95% CI and DF = 40, t* = 2.021 so CI is (13.3 – 12.4) =/- 2.021 x 0.544 which gives (- 0.20, 2.0)
c.) SRS of two normal independent pops, no outliers
78. xbarbeef = 155.3, st dev beef = 22.2 xbarpoultry = 122.5, st dev poultry = 25.5. Must pool.
a.) DF pooled = 20 + 17 – 2 = 35 so use 30 for t* = 2.042. SE = 7.93 so 95% CI is
(155.3 – 122.5) +/- 2.042 x 7.93 = ( 16.6, 49.0)
b.) zero is not included so difference could not be zero so reject
Chapter 12
2. a.) Response is likelihood they will buy, pops are those using each of three different water treatment devices, I = 3, ni is 75 each so N = 225
b.) Response is concrete strength, pops are concretes made with 5 different formulas, I = 5, ni is 6 each so N = 30
c..) Response is score, pops are students receiving a method, I = 3, ni is 20 each so N =60
4. a.) a.) DFG = 2, DFE = 222, DFT = 224, b.) DFG = 4, DFE = 25, DFT = 29, c.) DFG = 2, DFE = 57, DFT = 59
b.) a.) Ho: all types equally likely to be bought, Ha: all types not equally likely to be bought, b.) Ho: all mixtures equally strong, Ha: all mixtures not equally strong, c.) Ho: all methods equally effective, Ha: all methods not equally effective
c.) a.) Numerator DF = 2, denominator DF = 222, a.) Numerator DF = 4, denominator DF = 25, a.) Numerator DF = 2, denominator DF = 57.
1-way ANOVA project:
a.) sales
b.) Design with 4 levels
c.) unbiased estimate of mu4 is xbar4 which is 27.2
d.) unbiased estimate of the variance is either MSE or sp squared which is 10.5
e.) st dev assumption is checked by comparing the four s's or looking at Residuals vs Fitted Line and eyeballing that none are way higher/lower than another. Normal distr assumption is checked by looking at Normal Plot of residuals and insuring it is approx linear.
f.) Ho: all designs give the same sales. Ha: the designs do not give the same sales
test stat is F = 18.59. Its distr is F(3, 15) and the p-value is 0.0000 so I reject (p is less than any alpha)
Conclude in terms of prob: Sufficient evidence to believe the different designs do result in the different sales levels.
g.) It is appropriate to conduct follow on analysis only because we rejected that there is no difference.
1 is not stat diff from 2
1 is not stat diff from 3
1 is stat diff from 4
2 is not stat diff from 3
2 is stat diff from 4
3 is stat diff from 4
Tukey line test results in 4 being different from the rest and higher so it is better.
Chapter 13
2. a.) Response is hours of sleep, Factor A is type smoker, Factor B is gender, I = 3, J = 2, N = 160x3=480, b.) Response is concrete strength, Factor A is formula, Factor B is cycle, I = 6, J = 3, N = 6x9 = 54, c.) Response is score, Factor A is method, Factor B is major, I = 4, J = 2, N = 2x16 = 32.
4. a.) Sources of variation are Factor A, Factor B, Interaction of A*B and Error, DFA = 2, DFB = 1, DFAB = 2, DFE = 474, b.) Sources of variation are Factor A, Factor B, Interaction of A*B and Error, DFA = 5, DFB = 2, DFAB = 10, DFE = 38, c.) Sources of variation are Factor A, Factor B, Interaction of A*B and Error, DFA = 3, DFB = 1, DFAB = 3, DFE = 24.
12. You can plot so CS, EO and O are on the x-axis and get two lines, one for each gender or you can plot so M and F are on the x-axis and you get three lines, one for CS, for EO and for O. The plot shows the lines are not parallel so estimate there will be an interaction. Next look at the marginal means. Notice that there is an increase from male to female but there is an increase and decrease for major. You really can't look at a main effect here since there is an interaction and each main effect is not consistent. You could say for females the majors go up then down, for males the majors go steadily down.