2 Sample Mean Hypothesis t-test Example

Calculations by Professor S. Gramlich

Updated 3/22/09

 

9-3 #26 (from Triola, “Elementary Statistics,” 10th ed, C2006, p. 482)

Blanking Out on Tests

Many students have had unpleasant experience of panicking on a test because the first

question was exceptionally difficult.  The arrangement of test items was studied for its

effect on anxiety.  The following scores are measures of “debilitating test anxiety,”

which most of us call panic or blanking out (base on data from “Item Arrangement,

Cogntive Entry Characterstics, Sex and Test Anxiety as Predictors of Achievement

in Examination Performance,” by Klimko, Journal of Experimental Education,

Vol. 52, No. 4).  Is there sufficient evidence to support the claim that the

two populations of scores have the same mean? Is there sufficient evidence to support

the claim that the arrangement of the test items has an effect on the score?

 

Questions Arranged from Easy to Difficult Questions Arranged from Difficult to Easy

24.64                                                                           33.62

33.31                                                                           35.91

26.43                                                                           27.24

28.89                                                                           27.62

25.49                                                                           34.02

39.29                                                                           26.68

20.6                                                                             32.34

24.23                                                                           42.91

28.71                                                                           26.63

38.81                                                                           29.49

16.32                                                                           29.34

21.13                                                                           30.2

7.1                                                                               30.26

31.73                                                                           35.32

27.85                                                                           33.53

32.83                                                                           32.54

26.69  

32.86  

30.02  

30.29  

28.02  

28.9    

21.06  

21.96  

30.72

 

Step 1:  Hypotheses

H0:  μ1 = μ2  (original claim)     or      H0:  μ1 - μ2 = 0

H1:  μ1 ≠ μ2                                      H1:  μ1 - μ2 ≠ 0

(2 tailed test)

 

Step 2:  Given

assume: alpha = 5% = .05; observations independent, normal, srs; σ1 & σ2 unknown

 

SAMPLE 1 stats (n1 = 25)

sample mean:  xbar1 = Σx/n = (24.64+ 33.31+...+30.72)/ 25

or use excel "=AVERAGE(A2:A26)"

xbar1 = 27.115

 

sample standard deviation:

s1 = √[Σ(x-xbar)2 / (n-1)] = √[(24.64-27.115)2 + (33.31-27.115)2 +...+ (30.72-27.115)2] / (25-1)]

or use excel "=STDEV(A2:A26)"

s1 = 6.857

 

SAMPLE 2 stats (n2 = 16)

sample mean: xbar2 = Σx/n = (33.62 + 35.91 +...+ 32.54)/16

or use Excel "=AVERAGE(B2:B26)"

xbar2 = 31.728

 

sample standard deviation:

s2 = √[Σ(x-xbar)2 / (n-1)] = √[(33.62-31.728)2 + (35.91-31.728)2 +...+ (32.54-31.728)2/ (16-1)]

or use Excel "=STDEV(B2:B26)"

s2 = 4.26

 

Step 3:  t critical value (cv)

use df smaller of df1 = n1-1 = 25-1 =24 or df2 = n2-1 = 16-1 =15 on Table A-3

so with df = 15 and tc= ±2.131 (2 tail)

or use Excel “=TINV(0.05,15)”

 

Step 4:  t test statistic (ts)

t = [(xbar1 - xbar2) - (μ1 - μ2)] / √[ s12/n1 + s22/n2]

  = [(27.115 - 31.728) - 0] / √[6.8572/25 + 4.262/16]

  = -4.613/ √[47.018/25 + 18.148/16]

  = -4.613/ √[1.881 + 1.134]

  = -4.613/ √[3.015]

  = -4.613/ 1.736

t = -2.657

 

 

Step 5:  Draw t curve, label cv, ts, & shade critical region

critical region in red divided into 2 tails

p-value in blue

           -2.657   -2.131         +2.131

 

 

Step 6:  Traditional DR

Since t = -2.657 is inside critical region, Reject H0.

 

Step 7:  P-Value

From Table A-3, with df = 15, t = -2.657 falls in between 2.947 & 2.602,

so 0.01 < P-val < 0.02

or use Excel “=TDIST(2.657,15,2)”

 

Step 8:  P-Value DR

Since (0.01 < P-val < 0.02) < (α = .05), Reject H0.

 

Step 9:  Confidence Interval (CI)

E = tc * √[ s12/n1 + s22/n2]

   = 2.131 * 1.736 (from step 3 & 4)

   = 3.699

 

μ1 - μ2  = (xbar1 - xbar2) ± E

             = -4.613 ± 3.699

(xbar1 - xbar2) - E < μ1 - μ2 < (xbar1 - xbar2) + E

-4.613 - 3.699 < μ1 - μ2 < -4.613 + 3.699

-8.312 < μ1 - μ2 < -0.914

 

Step 10:  CI DR

Since CI does not contain 0, Reject H0.

 

Step 11:  Conclusion

There is not sufficient evidence to support the claim that the two populations have the same mean anxiety score.  It appears that the arrangement of  test items does have an effect on the score.