Question

Multiple myeloma, or blood plasma cancer, is characterized by increased blood vessel formulation (angiogenesis) in the bone marrow that is a predictive factor in survival. One treatment approach used for multiple myeloma is stem cell transplantation with the patient's own stem cells. The data show below represents the bone marrow microvessel density for patients who has a complete response to the stem cell transplant (as measured by blood and urine tests). The measurements were taken immediately prior to the stem cell transplant and at the time the complete response was determined. Patient Before After 1 158 284 2 186 214 3 202 101 4 353 227 5 416 290 6 426 176 7 441 290

(a) At the 0.05 level of significance, is there evidence that the mean bone marrow microvessel density is higher before the stem cell transplant than after the stem cell transplant? (Be sure to include the null and alternate hypotheses as well as the value of the test statistic. State the decision rule used and write you conclusion in a complete sentence.)

(b) Interpret the meaning of the p-value in (a).

(c) Construct and interpret a 95% confidence interval estimate of the mean difference in bone marrow microvessel density before and after the stem cell transplant.

(d) What assumption is necessary about the population distribution in order to perform the test in (a)?

Answer 1

a)

Ho :   µd=   0
Ha :   µd >   0

SAMPLE 1	SAMPLE 2	difference , Di =sample1-sample2	(Di - Dbar)²
158	284	-126.000	44822.939
186	214	-28.000	12930.939
202	101	101.000	233.653
353	227	126.000	1622.939
416	290	126.000	1622.939
426	176	250.000	26989.796
441	290	151.000	4262.224

	sample 1	sample 2	Di	(Di - Dbar)²
sum =	2182	1582.00	600.000	92485.429

mean of difference ,    D̅ =ΣDi / n =   85.714                  
                          
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    124.1541                  
                          
std error , SE = Sd / √n =    124.1541   / √   7   =   46.9258      
                          
t-statistic = (D̅ - µd)/SE = (   85.71428571   -   0   ) /    46.9258   =   1.8266
                          
Degree of freedom, DF=   n - 1 =    6                  
t-critical value , t* =        1.943   [excel function: =t.inv(α,df) ]               
Decision: test stat < criitcal value , Do not reject null hypothesis

b)

p-value =        0.0588   [excel function: =t.dist.rt(t-stat,df) ]

c)

sample size ,    n =    7          
Degree of freedom, DF=   n - 1 =    6   and α =    0.05  
t-critical value =    t α/2,df =    2.4469   [excel function: =t.inv.2t(α/2,df) ]      
                  
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    124.1541          
                  
std error , SE = Sd / √n =    124.1541   / √   7   =   46.9258
margin of error, E = t*SE =    2.4469   *   46.9258   =   114.8234
                  
mean of difference ,    D̅ =   85.714          
confidence interval is                   
Interval Lower Limit= D̅ - E =   85.714   -   114.8234   =   -29.109
Interval Upper Limit= D̅ + E =   85.714   +   114.8234   =   200.538
                  
so, confidence interval is (   -29.1091   < µd <   200.5377   )  

d)

b) sample are random and independent.

we assume that d has normal distribution