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