# Practice Final Summer 2013 KEY

```1)
The probability that this confidence interval captured the true average difference is 0 or 1
90% of all confidence intervals done this way would correctly capture the true average difference
We cannot show with 90% confidence that the true averages are statistically different
We are 90% confident the true average difference is between -1.47 and 5.83
There’s a 90% probability the next confidence interval captures the true average difference
2) H0: &micro;d ≥ 0
Ha: &micro;d &lt; 0
α = 0.05
t=(37.77-38.44)/(2.39/sqrt(49)) = -1.96
0.025 &lt; p-value &lt; 0.05
Reject
Our data shows the sun is lowering the hardness of the bricks
3)
142  142 
1 

142
200  200 
 1.555
=(0.66, 0.76)
200
200
4)
0.005=2.576s * sqrt( .5*(1-.5)/n )
N=(2.576/0.005)2*.5*(1-.5) = 66.357.76
Need 66358 people
5)
H0: &micro;=16
Ha: &micro;≠16
α = (somewhere between 0.08 to 0.15)
t=2.2 (accept 2 to 2.5)
pvalue should be around 0.014x2 = 0.03
Reject
The average temperature for a CSU student to stay home is significantly different from 16
6) Cannot do
7) 150 +- 2.398 * 25/sqrt(10) = (131.04, 168.95)
8) H0: Time is independent of light running
Ha: Time is related to light running
α = 0.05
17.67 20.32
22.32 25.67
1.06
0.918
0.836 0.730
z
22 40  16 46 0
38  38  38  38 
1  
1  
86  86  86  86 

40
46
 1.88
0.05 &lt; p-value &lt; 0.10
p-value =2*0.0301 = 0.0602
Fail to Reject
We cannot show the time of day affects whether people run the light
9) 1.183 +- 1.99 * 0.116974 = (1.597, 2.062)
We are 95% confident that for every dollar increase an engineering parent has in starting salary the
child engineer’s starting salary will go up by 1.597 to 2.062 dollars.
Bonus:
Groups
Error
Total
DF
SS
MS
F
4
15
19
24
30
54
6
2
3
```