Question

In: Statistics and Probability

Children in elementary schools in a US city were given two versions of the same test,...

Children in elementary schools in a US city were given two versions of the
same test, but with the order of questions arranged from easier to more
difficult in version A and in reverse order in version B. A randomly selected
group of 44 students were given version A; their mean grade was 83 with
standard deviation 5.6. Another randomly selected group of 38 students were
given version B; their mean grade was 81 with standard deviation 5.3.
A hypothesis test is to be performed to investigate whether the population
mean grade of students answering version A is the same as the population
mean grade of students answering version B.
(a) Name the hypothesis test that is appropriate to use in this situation.
(b) Using appropriate notation, which you should define, specify the null
and alternative hypotheses associated with the test.
(c) Calculate, by hand, the value of the estimated standard error of the
difference between the sample means.
(d) Calculate, by hand, the value of the test statistic.
(e) The sample sizes are both greater than 25, so you can assume that when
the null hypothesis is true, the test statistic follows (approximately) the
standard normal distribution. Complete the hypothesis test, carefully
detailing the conclusions of the test.

Solutions

Expert Solution

a)

2 Sample t test independent

b)

µ1 = mean grade of students answering version A

µ2 =mean grade of students answering version B

Ho :   µ1 - µ2 =   0
Ha :   µ1-µ2 ╪   0

c)

Sample #1   ---->   A
mean of sample 1,    x̅1=   83.00                  
standard deviation of sample 1,   s1 =    5.60                  
size of sample 1,    n1=   44                  
                          
Sample #2   ---->   B                  
mean of sample 2,    x̅2=   81.00                  
standard deviation of sample 2,   s2 =    5.30                  
size of sample 2,    n2=   38                  
                          
difference in sample means =    x̅1-x̅2 =    83.0000   -   81.0   =   2.00  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    5.4633                  
std error , SE =    Sp*√(1/n1+1/n2) =    1.2099                  
      

d)


t-statistic = ((x̅1-x̅2)-µd)/SE = (   2.0000   -   0   ) /    1.21   =   1.653

e)

Level of Significance ,    α =    0.05

Degree of freedom, DF=   n1+n2-2 =    80                  
  
p-value =        0.102240   (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value>α , Do not reject null hypothesis      

               
There is insufficient evidence to conclude that mean grade of students answering version A is different then the population mean grade of students answering version B.

Thanks in advance!

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