Question

A student wants to investigate the effects of real vs. substitute eggs on his favorite brownie recipe. Ten of his friends have come up with a consensus for the ranking of each of 8 batches on a scale from 1 to​ 10, and that has been recorded. Four of the batches were made with real​ eggs, four with substitute eggs. The judges tasted the brownies in a random order. The mean score for the real eggs was 6.76 with a standard deviation of 0.653. The mean score for the substitute eggs was 4.73 with a standard deviation of 0.393.

RealSubstitute45678EggsScore

Two vertical boxplots are shown side-by-side in a graph with horizontal axis labeled "Eggs" with categories "Real" and "Substitute" and vertical axis labeled "Score" from less than 4 to 8 plus in intervals of 1. The five horizontal segments of each boxplot have vertical coordinates as follows, listed here from bottom to top: "Real", minimum 6, first quartile 6.2, median 6.8, third quartile 7.4, maximum 7.6; "Substitute", minimum 4.1, first quartile 4.3, median 4.8, third quartile 5, maximum 5.1. All values are approximate.

ANOVA (Analysis of Variance)

Source	DF	Sum of Squares	Mean Square	​F-ratio	​P-value
Eggs	1	8.429470	8.42947	28.3782	0.0018
Error	6	1.782255	0.29704
Total	7	10.211725

​b) What can be concluded from the ANOVA​ table?

Identify the test statistic.

F = _ ? (Type an integer or a​ decimal.)

Identify the​ P-value.

P-value = _ ? (Type an integer or a​ decimal.)

Draw a conclusion for the test. Use a level of significance of α=0.05.

(Reject/ Fail to reject) the null hypothesis. The real eggs (have/do not have) a significantly different mean score.

​d) Perform a​ two-sample pooled​ t-test of the difference. What is the​ P-value? Compare the square of the​ t-statistic to the​ F-ratio.

P-value = _ ? (Round to three decimal places as​ needed.)

Answer 1

from anova:

F =28.3782

p value = 0.0018

p value < 0.05 , reject Ho

The real eggs (have) a significantly different mean score.

.........

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   6.76                  
standard deviation of sample 1,   s1 =    0.65                  
size of sample 1,    n1=   4                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   4.73                  
standard deviation of sample 2,   s2 =    0.39                  
size of sample 2,    n2=   4                  
                          
difference in sample means =    x̅1-x̅2 =    6.7600   -   4.7   =   2.03  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    0.5389                  
std error , SE =    Sp*√(1/n1+1/n2) =    0.3811                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   2.0300   -   0   ) /    0.38   =   5.327
                          
Degree of freedom, DF=   n1+n2-2 =    6                  
  
p-value =        0.002 (excel function: =T.DIST.2T(t stat,df) )         

(t stat)^2   = (5.327)^2 =28.378

F stat = 28.378

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

Please revert back in case of any doubt.

Please upvote. Thanks in advance.