## STAT 303, 503-504, Spring 1996 Prof. H. Joseph Newton Exam 2 Results

1. (#13 on form B) 0, 0, n-1,1
2. (#14) b
3. (#1) p=(100(.53)+150(.48))/250=0.5
4. (#2)
1. 3
2. pi=proportion of all kids in school who are left handed
3. H_0: pi=.2
4. H_a: pi not =.2
5. .826
6. .409
7. No, not enough evidence to conclude proportion of school being left handed different from 0.2
5. (#4)
1. 2
2. sigma_1^2/sigma_2^2=ratio of male to female variances
3. H_0: sigma_1^2/sigma_2^2 greater or equal to 1
4. H_a: sigma_1^2/sigma_2^2 < 1
5. .727
6. .181
7. No, cannot reasonably conclude female IQ's more variable
6. (#5)
1. 4
2. mu_d=mean of differences in times after and before drug
3. H_0: mu_d greater or equal 0
4. H_a: mu_d < 0
5. -3.333
6. .001
7. Yes, drug decreases time through maze on average
7. (#3)
1. 1
2. pi_1-pi_2=difference of proportions of men and women favoring Clinton
3. H_0: pi_1-pi_2 greater or equal to 0
4. H_a: pi_1-pi_2 < 0
5. -1.414
6. .079
7. No, not enough evidence to say there's a gender gap
8. (#6) Saying there's a gender gap when there isn't
9. (#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.
10. (#11) a) is A and C, b) is C and B, c) is C
11. (#8) F_{.40,10,10} is bigger
12. (#9) 0.5
13. (#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)
14. (#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. (#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. (#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.

### Results

```
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

```

### The Curve!

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

### Regressing the Second Exam Scores on the First

```
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
------------------------------------------------------------------------------
```