Question

If we increase our food intake, we generally gain weight. Nutrition scientists can calculate the amount of weight gain that would be associated with a given increase in calories. In one study, 16 nonobese adults, aged 25 to 36 years, were fed 1000 calories per day in excess of the calories needed to maintain a stable body weight. The subjects maintained this diet for 8 weeks, so they consumed a total of 56,000 extra calories. According to theory, 3500 extra calories will translate into a weight gain of 1 pound. Therefore, we expect each of these subjects to gain 56,000/3500 = 16 pounds (lb). Here are the weights before and after the 8-week period, expressed in kilograms (kg).

Subject	1	2	3	4	5	6	7	8
Weight before	55.7	54.9	59.6	62.3	74.2	75.6	70.7	53.3
Weight after	61.7	58.7	66.1	66.2	79.1	82.4	74.3	59.3
Subject	9	10	11	12	13	14	15	16
Weight before	73.3	63.4	68.1	73.7	91.7	55.9	61.7	57.8
Weight after	79.1	65.9	73.3	76.8	93.2	63.0	68.3	60.2

(a) For each subject, subtract the weight before from the weight after to determine the weight change in kg.

Subject 1		kg
Subject 2		kg
Subject 3		kg
Subject 4		kg
Subject 5		kg
Subject 6		kg
Subject 7		kg
Subject 8		kg
Subject 9		kg
Subject 10		kg
Subject 11		kg
Subject 12		kg
Subject 13		kg
Subject 14		kg
Subject 15		kg
Subject 16		kg

(b) Find the mean and the standard deviation for the weight change. (Round your answers to four decimal places.)

xkg	=	kg
skg	=	kg

(c) Calculate the standard error se and the margin of error me for 95% confidence. (Round your answers to four decimal places.)

se	=
me	=

Report the 95% confidence interval for weight change in a sentence that explains the meaning of the 95%. (Round your answers to four decimal places.)

Based on a method that gives correct results 95% of the time, the mean weight change is   kg to   kg.

(d) Convert the mean weight gain and standard deviation in kilograms to pounds. Because there are 2.2 kg per pound, multiply the value in kilograms by 2.2 to obtain pounds. (Round your answer to four decimal places.)

xlb	=	lb
slb	=	lb

Do the same for the confidence interval. (Round your answers to four decimal places.)

  ,

lb

(e) Test the null hypothesis that the mean weight gain is 16 lb. (Let

α = 0.05.)

Specify the null hypothesis.

H₀: μ > 16

H₀: μ ≠ 16

    

H₀: μ ≥ 16

H₀: μ = 16

H₀: μ ≤ 16

Specify the alternate hypothesis.

H_a: μ = 16

H_a: μ < 16

    

H_a: μ ≥ 16

H_a: μ ≠ 16

H_a: μ > 16

Carry out the test. (Round your answer for t to three decimal places.)

t =

Give the degrees of freedom.


Give the P-value. (Round your answer to four decimal places.)

Answer 1

a)

SAMPLE 1	SAMPLE 2	difference , Di =sample1-sample2
61.7	55.7	6.000
58.7	54.9	3.800
66.1	59.6	6.500
66.2	62.3	3.900
79.1	74.2	4.900
82.4	75.6	6.800
74.3	70.7	3.600
59.3	53.3	6.000
79.1	73.3	5.800
65.9	63.4	2.500
73.3	68.1	5.200
76.8	73.7	3.100
93.2	91.7	1.500
63	55.9	7.100
68.3	61.7	6.600
60.2	57.8	2.400

b)

mean of difference ,    D̅ =ΣDi / n =   4.7312
      
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    1.7790

c)

std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    1.7790          
                  
std error , SE = Sd / √n =    1.7790   / √   16   =   0.4448

confidence interval is                   
Interval Lower Limit= D̅ - E =   4.731   -   0.9480   =   3.7833
Interval Upper Limit= D̅ + E =   4.731   +   0.9480   =   5.6792

d)

mean of difference ,    D̅ = 10.4088
              
std dev of difference , Sd =   3.9139

confidence interval is                       
Interval Lower Limit= D̅ - E =      8.3232
Interval Upper Limit= D̅ + E =   12.4943

e)

Ho :   µd=   16
Ha :   µd <   16

mean of difference ,    D̅ =ΣDi / n =   4.7312       10.4088          
               3.9139          
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    1.7790                  
                          
std error , SE = Sd / √n =    1.7790   / √   16   =   0.4448      
                          
t-statistic = (D̅ - µd)/SE = (   4.73125   -   16   ) /    0.4448   =   -25.337
                          
Degree of freedom, DF=   n - 1 =    15                  
t-critical value , t* =        -1.7531   [excel function: =t.inv(α,df) ]              
                          
p-value =        0.0000   [excel function: =t.dist(t-stat,df) ]              
Decision:   p-value <α , Reject null hypothesis