Question

The population mean for a statistics test in a class that is usually taken by juniors and seniors is 75 (out of 85). A random sample of 15 sophomores is selected from the course. The research question is whether or not sophomores are different from the average student. The scores are: 76,76,66,49,84,72,72,50,56,62,65,71,60,82,73

A) State the null and alternative hypothesis: which I have null being equal to 75 and alternative being not equal to 75.

B) Explain why T-Test is suitable.

My answer: QQ Plot is linear, and the population size is small?

C) Using the test described in part b and the significance level of .05, compute the observed test statistics. (What do they mean by test statistics too)

D) Draw a conclusion in the context of the problem based on the results from part c using the rejection region method

E) Computer a 95% confidence interval for the sophomores and draw a conclusion in the context of the problem based on the interval

Answer 1

A) Since we have to test if sophomores are different from average students,        
the null and alternative hypothesis are        
Ho : μ = 75       
Ha : μ ≠ 75              where μ is the population mean     
        
B) T-test is suitable because of following reasons       
1)   Population standard deviation is not known       
2)   Sample size is small ( < 30)       
3)   QQ plot is linear, hence population is normal       
        
C) α = 0.05 5% level of significance      
From the given data we find the mean and standard deviation using        
Excel functions average and stdev.s       
x̅ = 67.6  Mean     
s = 10.6422  Standard Deviation     
Test statistic t is given by        
        
             
Observed Test statistic t = -2.6931       
        
D) Degrees of freedom = n - 1 = 15 - 1 = 14       
For α = 0.05 and degrees of freedom = 14       
we find the critical t for two sided test using Excel function t.inv.2t       
t-crit = T.INV.2T(0.05, 14)       
           = 2.1448       
Rejection Region is        
Reject Ho if |Observed test statistic t| > t-crit       
|Observed test statistic t| = |-2.6931| = 2.6931       
2.6931 > 2.1448       
Hence, we reject Ho       
Conclusion :       
There is enough statistical evidence to conclude that       
the sophomores are different from the average student       
        
E) Confidence interval for mean is given by       
            
For 95%, α = 0.05       
From Excel function T.INV.2T(0.05, 14)           (14 is the degrees of freedom)   
t = T.INV.2T(0.05, 14) = 2.1448                  (We take positive value)   
        
   = (61.7065, 73.4935)       
95% confidence interval for the sophomores is (61.7065, 73.4935)       
Since the population mean 75 is not a part of the above confidence interval,       
it is right that we reject Ho. That is, it is statistically proven that        
population mean μ ≠ 75