Question

A professor wants to know whether or not there is a difference in comprehension of a lab assignment among students depending on if the instructions are given all in text, or if they are given primarily with visual illustrations. She randomly divides her class into two groups of 15, gives one group instructions in text and the second group instructions with visual illustrations. The following data summarizes the scores the students received on a test given after the lab. Let the populations be normally distributed with a populations standard deviation of 5.32 points for both the text and visual illustrations.
Text (Group Visual Illustrations (Group 2) 57.3 59 45.3 57.6 87.1 72.9 61.2 83.2 43.1 64 87.3 76.7 75.2 78.2 88.2 64.4 67.5 89 86.2 72.9 67.2 88.2 54.4 43.8 93 97.1 89.2 95.1 52 84.1 Is there evidence to suggest that a difference exists in the comprehension of the lab based on the test scores? Use a=0.10. Enter the test statistic - round to 4 decimal places. Enter the P-Value - round to 4 decimal places. Can it be concluded that a difference exists in the comprehension of the lab based on the test scores?

Answer 1

Answer:

It is concluded that no difference exists in the comprehension of the lab based on the test scores.

Step-by-step explanation:

Let the populations be normally distributed with a populations standard deviation of 5.32 points for both the text and visual illustrations.

The above statement tells us that the independent t or two sample t test will be performed as both have equal variances and are normally distributed.

This can be easily done through excel.

The following table is obtained

     

t-Test: Two-Sample Assuming Equal Variances  

 

                 Text                      Visual Illustrations

Mean          70.28                       75.08

Variance 304.48                     228.58

Observations 15                             15

Pooled Variance Sp²= 266.53

       Pooled Standard Deviation = Sp = 16.33

Hypothesized Mean Difference = x1`-x2`= 0

df = n1+n2-2= 15+15-2= 30-2= 28

t Stat -0.805188239

P(T<=t) two-tail 0.427495979                      

t Critical two-tail 1.701130908                              

Let the null and alternate hypotheses be

H0 : u1-u2= 0   against the claim Ha: u1-u2≠0

There is no difference between the means

against the claim

that there is a difference in means of a lab assignment among students depending on if the instructions are given all in text, or if they are given primarily with visual illustrations.

The significance level is ∝= 0.1

The d.f is n1+n2-2= 15+15-2= 30-2=28

This is a two tailed test and the critical region is t (0.025) (28) ≥ 1.7011 and  t (0.025) (28) ≤ - 1.7011.

The test statistic is

t= x1-x2/ Sp √1/n1+ 1/n2

t= 70.28 -75.08/ 16.33√1/15 +1/15

t= -4.8/5.963

t= -0.8049811  ( minute difference from excel result due to rounding)

Since the calculated value of t= -0.8049 does not fall in the critical region we conclude that there is no difference in means of a lab assignment among students depending on if the instructions are given all in text, or if they are given primarily with visual illustrations.

We accept the null hypothesis.

The p- value is 0.427495979.