Question

1. In many animal species the males and females differ slightly in structure, coloring, and/or size. The hominid species Australopithecus is thought to have lived about 3.2 million years ago. (“Lucy,” the famous near complete skeleton discovered in 1974, is an Australopithecus .) Forensic anthropologists use partial skeletal remains to estimate the mass of an individual. The data below are estimates of masses from partial skeletal remains of this species found in sub-Saharan Africa. Appropriate graphical displays of the data indicate that it is reasonable to assume that the population distributions of mass are approximately normal for both males and females. You may also assume that these samples are representative of the respective populations. Estimates of mass (kg)

Males 51.0, 45.4, 45.6, 50.1, 41.3, 42.6, 40.2, 48.2, 38.4, 45.4, 40.7, 37.9, 41.3, 31.5

Females 27.1, 33.5, 28.0, 30.3, 32.7, 32.5, 34.2, 30.5, 27.5, 23.3,35.7

Do these data provide convincing evidence that the mean estimated masses differ for Australopithecus males and females? Provide appropriate statistical justification for your conclusion.

2. In an introductory marketing class students were presented with 6 items they could bid on in an auction. They were asked to bid privately and also estimate the “typical” bid for each item by their classmates. The items were randomly selected from a large list of items that students might purchase. An initial analysis of the data established the plausibility that the distribution of differences (estimated – actual) is approximately normal.

Construct a 95% confidence interval for the mean difference between the actual bid and the estimated “typical” bid for the population of items.

GOOD	ACTUAL	ESTIMATE	DIFFERENCE
Teddy bear	1.00	4.90	3.90
Music CD	1.25	4.53	3.28
sachet	2.70	5.44	2.74
wood puzzle	3.00	5.17	2.17
smoked salmon	3.00	6.67	3.67
jelly beans	4.00	7.30	3.30

Answer 1

1)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
  
Sample #1   ---->   1                  
mean of sample 1,    x̅1=   42.829                  
standard deviation of sample 1,   s1 =    5.246                  
size of sample 1,    n1=   14                  
                          
Sample #2   ---->   2                  
mean of sample 2,    x̅2=   30.482                  
standard deviation of sample 2,   s2 =    3.705                  
size of sample 2,    n2=   11                  
                          
difference in sample means =    x̅1-x̅2 =    42.8286   -   30.5   =   12.347  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    4.6395                  
std error , SE =    Sp*√(1/n1+1/n2) =    1.8693                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   12.3468   -   0   ) /    1.87   =   6.605
                          
Degree of freedom, DF=   n1+n2-2 =    23                  
p-value =        0.0000 (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value <α , Reject null hypothesis                      
data provide convincing evidence that the mean estimated masses differ for Australopithecus males and female

2)

SAMPLE 1	SAMPLE 2	difference , Di =sample1-sample2	(Di - Dbar)²
4.9	1	3.900	0.523
4.53	1.25	3.280	0.011
5.44	2.7	2.740	0.191
5.17	3	2.170	1.013
6.67	3	3.670	0.243
7.3	4	3.300	0.015

	sample 1	sample 2	Di	(Di - Dbar)²
sum =	34.01	14.95	19.060	1.997

sample size ,    n =    6          
Degree of freedom, DF=   n - 1 =    5   and α =    0.05  
t-critical value =    t α/2,df =    2.5706   [excel function: =t.inv.2t(α/2,df) ]      
                  
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    0.6319          
                  
std error , SE = Sd / √n =    0.6319   / √   6   =   0.2580
margin of error, E = t*SE =    2.5706   *   0.2580   =   0.6631
                  
mean of difference ,    D̅ =   3.177          
confidence interval is                   
Interval Lower Limit= D̅ - E =   3.177   -   0.6631   =   2.514
Interval Upper Limit= D̅ + E =   3.177   +   0.6631   =   3.840
                  
so, confidence interval is (   2.5135   < µd <   3.8398   )