Prof. H. Joseph Newton

Exam 2 Results

- (#13 on form B) 0, 0, n-1,1
- (#14) b
- (#1) p=(100(.53)+150(.48))/250=0.5
- (#2)
- 3
- pi=proportion of all kids in school who are left handed
- H_0: pi=.2
- H_a: pi not =.2
- .826
- .409
- No, not enough evidence to conclude proportion of school being left handed different from 0.2

- (#4)
- 2
- sigma_1^2/sigma_2^2=ratio of male to female variances
- H_0: sigma_1^2/sigma_2^2 greater or equal to 1
- H_a: sigma_1^2/sigma_2^2 < 1
- .727
- .181
- No, cannot reasonably conclude female IQ's more variable

- (#5)
- 4
- mu_d=mean of differences in times after and before drug
- H_0: mu_d greater or equal 0
- H_a: mu_d < 0
- -3.333
- .001
- Yes, drug decreases time through maze on average

- (#3)
- 1
- pi_1-pi_2=difference of proportions of men and women favoring Clinton
- H_0: pi_1-pi_2 greater or equal to 0
- H_a: pi_1-pi_2 < 0
- -1.414
- .079
- No, not enough evidence to say there's a gender gap

- (#6) Saying there's a gender gap when there isn't
- (#7) If male and female IQ's have the same population variances, then the chance of observing sample variances this much different (200 versus 275) or more is 0.181, which is not unusual.
- (#11) a) is A and C, b) is C and B, c) is C
- (#8) F_{.40,10,10} is bigger
- (#9) 0.5
- (#10) b, d (Not (a) because a type I error would be to say that there's not much variability when there is which could lead to using defective O-rings which could cause shuttle to explode, not (c) because we use chi-square test)
- (#12) b, c (Not (a) because the point of the procedure is to analyze the sources of variability, not (d) because there is no one sided test in ANOVA)
- (#15) Any 2 of b, d, e (Not (a) because rejecting at 0.05 means p-value < 0.05 which is not necessarily < 0.01, not (c) because the probability of a Type I error is alpha no matter what the sample size is)
- (#16) Statistical theory tells us that if we obtained many random samples like the one we have and calculated the 95% CI for each one, then 95% of the CI's would have the true value of the parameter in them. Thus we're 95% confident that the particular random sample we have has resulted in a CI that contains the true value of the parameter.

Mean=63.15, Std. Dev=14.60 Q1=52, Q2=64, Q3=74, IQR=22 Stem-and-leaf plot for exam2 1. | 8 2* | 2. | 3* | 2 3. | 9 4* | 0224 4. | 7777899 5* | 0001223 5. | 56778889 6* | 022223334 6. | 5567888 7* | 00111133344 7. | 555555666 8* | 001114 8. | 9* | 11 9. | 10* | 0

A: 80-100 (9) B: 62-79 (35) C: 47-61 (23) D: 32-46 (6) F: <32 (1)

regress exam2 exam1 Source | SS df MS Number of obs = 74 ---------+------------------------------ F( 1, 72) = 60.36 Model | 7095.48888 1 7095.48888 Prob > F = 0.0000 Residual | 8463.87598 72 117.553833 R-squared = 0.4560 ---------+------------------------------ Adj R-squared = 0.4485 Total | 15559.3649 73 213.141984 Root MSE = 10.842 ------------------------------------------------------------------------------ exam2 | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- exam1 | .6295825 .0810364 7.769 0.000 .4680395 .7911256 _cons | 22.23429 5.414993 4.106 0.000 11.4397 33.02888 ------------------------------------------------------------------------------