Question

6.) In my CENG 102 course from 2015, the final was split over 2 separate days, with half the class in each exam section. A small random sample of scores from each exam section are provided below.

Wednesday Exam: 431, 392, 392, 372, 426, 441, 448, 452 (n = 8)

Friday Exam: 450, 375, 483, 391, 415, 460, 395, 446, 326, 332 (n = 10)

Assume the distributions of grades for each section are normally distributed (and for simplicity, the distributions are continuous).

a.) Provide a 2-sided 95% confidence interval for (i.) the Wednesday exam scores, (ii.) the Friday exam scores, and (iii.) the difference between the two scores.

b.) Students are concerned the exams had different levels of difficulty. A preliminary look at the two means suggests the first exam had a higher average. Test this claim. Can you conclude that the Wednesday exam had a higher average than the Friday exam?

c.) At 5% significance, determine the rejection region for the test.

d.) What is the power of the 5% level test in the difference between the two exam score samples, given that the actual difference was 26? What is the probability of a Type II error?

e.) What is the probability of a Type I error?

f.) If the probability of a Type II error is low, why? If the probability is high, what can be done to reduce the likelihood of this error?

g.) On a practical level, on an exam worth 500 points, is a difference of 26 points a cause for concern?

Caution about problem 6:

Problem 6 is meant to simply give you practice on hypothesis tests, error, and power – using real data. Technically, several assumptions allowing us to attempt this hypothesis test are violated. They are:

1.) Although the exam scores were approximately normally distributed, the population was not large enough. Doing a simple random sample would mean a moderate chance of selecting the same exam/score more than once. I did not do that. Once one score was picked, it was not used again – which makes the distribution hypergeometric in a sense (taking one “test” out of the population technically changes the remaining population parameters).

2.) Scores were discrete, not continuous – although with a large spread, this likely has negligible effects.

3.) For a proper “experiment,” the sample should be less than 1% of the population. This was nowhere near true in this problem.

Answer 1

a)

i)

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   7          
't value='   tα/2=   2.3646   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   29.9368   / √   8   =   10.584271
margin of error , E=t*SE =   2.3646   *   10.58427   =   25.027823
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    419.25   -   25.027823   =   394.222177
Interval Upper Limit = x̅ + E =    419.25   -   25.027823   =   444.277823
95%   confidence interval is (   394.22   < µ <   444.28   )
.....

ii)

friday

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   9          
't value='   tα/2=   2.2622   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   53.3626   / √   10   =   16.874735
margin of error , E=t*SE =   2.2622   *   16.87473   =   38.173302
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    407.30   -   38.173302   =   369.126698
Interval Upper Limit = x̅ + E =    407.30   -   38.173302   =   445.473302
95%   confidence interval is (   369.13   < µ <   445.47   )

......

iii)

Degree of freedom, DF=       14          
t-critical value =    t α/2 =    2.145   (excel formula =t.inv(α/2,df)      
                  
                  
                  
std error , SE =    √(s1²/n1+s2²/n2) =    19.919          
margin of error, E = t*SE =    2.145   *   19.919   =   42.7229
                  
difference of means = x̅1-x̅2 =    419.2500   -   407.300   =   11.9500
confidence interval is                   
Interval Lower Limit = (x̅1-x̅2) - E =    11.9500   -   42.723   =   -30.7729
Interval Upper Limit = (x̅1-x̅2) + E =    11.9500   -   42.723   =   54.6729

...................

b)

Ho :   µ1 - µ2 =   0          
Ha :   µ1-µ2 >   0          
                  
Level of Significance ,    α =    0.05          
                  
Sample #1   ---->   sample 1          
mean of sample 1,    x̅1=   419.25          
standard deviation of sample 1,   s1 =    29.93683827          
size of sample 1,    n1=   8          
                  
Sample #2   ---->   sample 2          
mean of sample 2,    x̅2=   407.300          
standard deviation of sample 2,   s2 =    53.36          
size of sample 2,    n2=   10          
                  
difference in sample means = x̅1-x̅2 =    419.250   -   407.3000   =   11.9500
                  
std error , SE =    √(s1²/n1+s2²/n2) =    19.9194          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   11.9500   /   19.9194   ) =   0.5999
                  
Degree of freedom, =   14          
                  
                     
p-value =        0.2791   [excel function: =T.DIST.RT(t stat,df) ]      
Conclusion:     p-value>α , Do not reject null hypothesis              

no, we can not conclude that the Wednesday exam had a higher average than the Friday exam

...............

c)

t-critical value , t* =        1.7459   (excel function: =t.inv(α,df)          

critical region : test stat > 1.7459   
here,

Decision:   | t-stat | < | critical value |, so, Do not Reject Ho          

...............

d)

std error , SE =    √(s1²/n1+s2²/n2) =    19.9194
(x - µo)/SE =  Zα/2

Zα/2   = 1.645   (RIGHT tailed test)

SO,

x= Zα/2 *SE + µo

= 1.645*19.9194 + 11.95

x =44.72

ß= P ( Z < =(x̄-true mean)/SE )

= P(Z<=0.9398)

=0.8263

POWER = 1-ß = 1- 0.8263

= 0.1737

e)

probability of type 1 error = 0.05

f)

PROBABILITY of type II error is very high.

by ensuring your test has enough power, you can reduce error

sample size should large enough.

increase the level of significance

please revert back for doubt