Question

1.Child Health and Development Studies (CHDS) has been collecting data about expectant mothers in Oakland, CA since 1959. One of the measurements taken by CHDS is the weight increase (in pounds) for expectant mothers in the second trimester.

In a fictitious study, suppose that CHDS finds the average weight increase in the second trimester is 14 pounds. Suppose also that, in 2015, a random sample of 42 expectant mothers have mean weight increase of 15.7 pounds in the second trimester, with a standard deviation of 6.0 pounds.

A hypothesis test is done to see if there is evidence that weight increase in the second trimester is greater than 14 pounds.

Find the p-value for the hypothesis test.

2.

According to Facebook’s self-reported statistics, the average Facebook user has 130 Facebook friends. For a statistics project a student at Contra Costa College (CCC) tests the hypothesis that CCC students will average more than 130 Facebook friends. She randomly selects 3 classes from the schedule of classes and distributes a survey in these classes. Her sample contains 45 students.

From her survey data she calculates that the mean number of Facebook friends for her sample is: ¯x= 138.7 with a standard deviation of: s=79.3.

She chooses a 5% level of significance. What can she conclude from her data?

1. Nothing. The conditions for use of a t-model are not met. She cannot trust that the p-value is accurate for this reason.
2. We cannot conclude that the average number of Facebook friends for CCC students is greater than 130. The sample mean of 138.7 is not significantly greater than 130.
3. Her data supports her claim. The average number of Facebook friends for CCC students is significantly greater than 130.

Answer 1

1)

Ho :   µ =   14                  
Ha :   µ >   14       (Right tail test)          
                          
Level of Significance ,    α =    0.05                  
sample std dev ,    s =    6.0000                  
Sample Size ,   n =    42                  
Sample Mean,    x̅ =   15.7000                  
                          
degree of freedom=   DF=n-1=   41                  
                          
Standard Error , SE = s/√n =   6.0000   / √    42   =   0.9258      
t-test statistic= (x̅ - µ )/SE = (   15.700   -   14   ) /    0.9258   =   1.836
                          
  
p-Value   =   0.036793   [Excel formula =t.dist.rt(t-stat,df) ]              

2)

Ho :   µ =   130                  
Ha :   µ >   130       (Right tail test)          
                          
Level of Significance ,    α =    0.05                  
sample std dev ,    s =    79.3000                  
Sample Size ,   n =    45                  
Sample Mean,    x̅ =   138.7000                  
                          
degree of freedom=   DF=n-1=   44                  
                          
Standard Error , SE = s/√n =   79.3000   / √    45   =   11.8213      
t-test statistic= (x̅ - µ )/SE = (   138.700   -   130   ) /    11.8213   =   0.736
                          
  
p-Value   =   0.232831   [Excel formula =t.dist(t-stat,df) ]              
Decision:   p-value>α, Do not reject null hypothesis                       

We cannot conclude that the average number of Facebook friends for CCC students is greater than 130. The sample mean of 138.7 is not significantly greater than 130.