Question

3. (60 pts.) A nurse in public health conducted a survey of women who had recently delivered babies.   A random sample of mothers who had 5 or fewer prenatal visits and a random sample of mothers who had 6 or more prenatal visits. The birthweights (in ounces) of each group are listed below. The birthweights of both groups can be considered to be normally distributed.

Less	More
49	133
108	108
110	93
82	119
93	119
114	98
134	106
114	131
96	87
52	153
101	116
114	129
120	97
116	110

1. At the 5% significance level, could the nurse conclude that the babies of mothers who had 6 or more prenatal visits had higher birthweights than those of mothers who had 5 or less prenatal visits?   

2. Construct a 90% confidence interval for difference between the mean weights of the two groups of babies.  

3. Interpret the confidence interval in terms of this problem.

4. What is the point estimate of this confidence interval?

5. What is the margin of error of this confidence interval?

Answer 1

a)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 <   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   100.21                  
standard deviation of sample 1,   s1 =    24.59                  
size of sample 1,    n1=   14                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   114.21                  
standard deviation of sample 2,   s2 =    18.15                  
size of sample 2,    n2=   14                  
                          
difference in sample means =    x̅1-x̅2 =    100.2143   -   114.2   =   -14.00  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    21.6073                  
std error , SE =    Sp*√(1/n1+1/n2) =    8.1668                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -14.0000   -   0   ) /    8.17   =   -1.714
                          
Degree of freedom, DF=   n1+n2-2 =    26                  
t-critical value , t* =        -1.7056   (excel function: =t.inv(α,df)              
Decision:   | t-stat | > | critical value |, so, Reject Ho                      
p-value =        0.049190   [ excel function: =T.DIST(t stat,df) ]               
Conclusion:     p-value <α , Reject null hypothesis                      

There is not suffiecient evidence to conclude that the babies of mothers who had 6 or more prenatal visits had higher birthweights than those of mothers who had 5 or less prenatal visits

b)

Degree of freedom, DF=   n1+n2-2 =    26              
t-critical value =    t α/2 =    1.7056   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    21.6073              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    8.1668              
margin of error, E = t*SE =    1.7056   *   8.1668   =   13.9294  
                      
difference of means =    x̅1-x̅2 =    100.2143   -   114.214   =   -14.0000
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    -14.0000   -   13.9294   =   -27.9294
Interval Upper Limit=   (x̅1-x̅2) + E =    -14.0000   +   13.9294   =   -0.0706

c)

We can interpret that 90% confident that difference of mean is between these two interval. Hence can say that 90 % condident that conclude that the babies of mothers who had 6 or more prenatal visits had higher birthweights than those of mothers who had 5 or less prenatal visits

d)

difference of means =    x̅1-x̅2 =    100.2143   -   114.214   =   -14.0000

e)

margin of error, E = t*SE =    1.7056   *   8.1668   =   13.9294

Please revert back in case of any doubt.

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