Question

In: Statistics and Probability

The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures the motivation,...

The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures the motivation, attitude toward school, and study habits of students. Scores range from 0 to 200. A selective private college gives the SSHA to an SRS of both male and female first-year students. The data for the women are as follows: 154 109 137 115 152 140 154 178 101 103 126 126 137 165 165 129 200 148 Here are the scores of the men: 108 140 114 91 180 115 126 92 169 146 109 132 75 88 113 151 70 115 187 104

Most studies have found that the mean SSHA score for men is lower than the mean score in a comparable group of women. Is this true for first-year students at this college? Use a 1% significance level.

(a) Hypotheses and results:

(b) Draw a picture and label p-value and horizontal axis:

(c) Draw a conclusion. Don’t just accept or reject. Say what it means in terms of this problem.

Solutions

Expert Solution

a)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 >   0                  
                          
Level of Significance ,    α =    0.01                  
                          
Sample #1   ---->   Women   
mean of sample 1,    x̅1=   141.06                  
standard deviation of sample 1,   s1 =    26.44                  
size of sample 1,    n1=   18                  
                          
Sample #2   ---->   Men
mean of sample 2,    x̅2=   121.25                  
standard deviation of sample 2,   s2 =    32.85                  
size of sample 2,    n2=   20                  
                          
difference in sample means =    x̅1-x̅2 =    141.0556   -   121.3   =   19.81  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    29.9938                  
std error , SE =    Sp*√(1/n1+1/n2) =    9.7448                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   19.8056   -   0   ) /    9.74   =   2.032
                          
Degree of freedom, DF=   n1+n2-2 =    36                  

b)

  
p-value =        0.024769   [excel function: =T.DIST.RT(t stat,df) ]      

c)


Conclusion:     p-value>α , Do not reject null hypothesis                      
                          
There is not enough evidence to support that  the mean SSHA score for men is lower than the mean score in a comparable group of women.

Please revert in case of any doubt.

Please upvote. Thanks in advance


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