1 n 16




2 mean 8.96 8.9625



3 stdev 0.38 0.3757



4 p(ѵ ≤ 8.5) = 0.11 0.1092



5 p(ѵ ≥ 10.0) 0.0029 0.0029



6 p(ѵ ≤ )=0.20 8.65 8.6463



7 stand error 0.09 0.0939



8 tc 2.13 2.1314



9 E 0.2 0.2002



10 p( 8.76 9.16 ) = 0.95


p( 8.7623 ≤ µ ≤ 9.1627 ) = 0.95

11 inc 8.73? no 8.73 is NOT a possible pop mean for the given data



12
faster












Homework 10.2 due 09 April 2010:






Write the null hypothesis






Write the alternate hypothesis






Use alpha of 0.05






Calculate tc






Calculate t-statistic t






Do you “reject the null hypothesis” OR “fail to reject the null hypotheis”?














Dist Speed

Speed
Speed (kph)
10.3 8.8

8.25 <tr><td class="n"> 8.2 </td></tr>
10.3 8.77

8.52 <tr><td class="n"> 8.5 </td></tr>
10.4 9.19

8.67 <tr><td class="n"> 8.7 </td></tr>
10.4 9.68

8.68 <tr><td class="n"> 8.7 </td></tr>
10.4 8.89

8.7 <tr><td class="n"> 8.7 </td></tr>
10.4 9

8.77 <tr><td class="n"> 8.8 </td></tr>
10.5 8.68

8.8 <tr><td class="n"> 8.8 </td></tr>
10.6 9.44

8.89 <tr><td class="n"> 8.9 </td></tr>
10.9 8.67

9 <tr><td class="n"> 9 </td></tr>
11.5 8.25

9.09 <tr><td class="n"> 9.1 </td></tr>
11.5 8.7

9.1 <tr><td class="n"> 9.1 </td></tr>
11.5 9.1

9.19 <tr><td class="n"> 9.2 </td></tr>
11.5 9.3

9.27 <tr><td class="n"> 9.3 </td></tr>
11.5 9.27

9.3 <tr><td class="n"> 9.3 </td></tr>
12.8 9.09

9.44 <tr><td class="n"> 9.4 </td></tr>
13 8.52

9.68 <tr><td class="n"> 9.7 </td></tr>








March 2009

skew 0.08
-0.04
Dist Speed
kurt -0.15
0.1
10.44 8.44
mode

8.7
11.5 8.82
median 8.95
8.95
12.69 8.67
mean 9.04
9.05
13.48 9.24
stdev 0.37
0.38
16.7 8.46



8.67
mean: 8.73



9.8



min 8.25
8.2



max 9.68
9.7



range 1.43
1.5



classes 5
5



width 0.29
0.3



cul cul f f rnd



8.53 8.5 2 2



8.82 8.8 5 5



9.11 9.1 4 4



9.39 9.4 3 4



9.68 9.7 2 1





16 16