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Chem 55 Exam 2 Solutions

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CHEM <strong>55</strong> Fall 2007 <strong>Exam</strong> 1 Copy # 1<br />

Name: Au,Hai Nguyen<br />

Values of Student's t<br />

% confidence interval<br />

°freedom 50 80 95 99<br />

1 1.00 3.08 12.71 63.66<br />

2 0.82 1.89 4.30 9.92<br />

3 0.76 1.64 3.18 5.84<br />

4 0.74 1.53 2.78 4.60<br />

5 0.73 1.48 2.57 4.03<br />

6 0.72 1.44 2.45 3.71<br />

7 0.71 1.41 2.36 3.50<br />

8 0.71 1.40 2.31 3.36<br />

9 0.70 1.38 2.26 3.25<br />

10 0.70 1.37 2.23 3.17<br />

20 0.69 1.33 2.09 2.85<br />

50 0.68 1.30 2.01 2.68<br />

100 0.68 1.29 1.98 2.63<br />

∞ 0.67 1.28 1.96 2.58<br />

Values of Q for rejection of data:<br />

Q<br />

90% confidence<br />

Number of<br />

Observations<br />

0.76 4<br />

0.64 5<br />

0.56 6<br />

0.51 7<br />

0.47 8<br />

0.44 9<br />

0.41 10<br />

Average, Sample standard deviation<br />

and confidence interval formulae:<br />

x =<br />

n<br />

∑<br />

i=<br />

1<br />

n<br />

n<br />

x<br />

∑<br />

( xi<br />

− x)<br />

i=<br />

1<br />

s =<br />

n −1<br />

CI = x ±<br />

ts<br />

n<br />

( x − μ)<br />

t =<br />

s<br />

i<br />

n<br />

2<br />

Error propagation formulae:<br />

for : y = x1<br />

+ x2<br />

for : y = x1⋅<br />

x2<br />

for : y = log( x)<br />

for : y = 10<br />

for : y = ln( x)<br />

for : y = e<br />

for : y = x<br />

x<br />

a<br />

x<br />

e<br />

e<br />

y<br />

y<br />

% e<br />

y<br />

e<br />

y<br />

% e<br />

y<br />

=<br />

=<br />

ex<br />

ey<br />

=<br />

x<br />

= y ⋅e<br />

x<br />

e<br />

2<br />

x1<br />

= a ⋅%<br />

e<br />

% e<br />

x<br />

+ e<br />

2<br />

x1<br />

ex<br />

ey<br />

= 0.434<br />

x<br />

= 2.303⋅<br />

y ⋅e<br />

x<br />

2<br />

x2<br />

+ % e<br />

2<br />

x2

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