A psychologist theorized that people can hear better when they have just eaten a large meal. Six individuals were randomly assigned to eat either a large meal or a small meal. After eating the meal, their hearing was tested. The hearing ability scores (high numbers indicate greater ability) are given in the following table.
Group Subject Hearing
Big_meal A 21
Big_meal B 25
Big_meal C 24
Small_meal D 19
Small_meal E 23
Small_meal F 22
Using the .05 level, do the results support the psychologist's theory? Complete questions below.
Let Sample 1 be the big meal group and Sample 2 be the small meal group. Determine the research hypotheses.
1. The research hypothesis is that the population mean for the big meal group is ______.
Question 1 options:
A. the same as 

B. greater than 

C. less than 

D. different from 
Determine the null hypotheses.
2. The null hypothesis is that the population mean for the big meal group is ______ the population mean for the small meal group.
Question 2 options:
A. less than 

B. the same as or less than 

C. different from 

D. greater than 
3. Calculate the t score.
t= ____
4. Determine the cutoff score(s) on the comparison distribution for the t test.
5. Reach a decision. Should reject or not reject the null hypothesis?
Question 5 options:
A. Reject the null hypothesis 

B. Do not reject the null hypothesis 
6. There is _____ evidence that there is a difference between the means.
Question 6 options:
A. sufficient 

B. insufficient 
7. Explain your answers to someone who has never had a course in statistics. The results _______ the psychologist's theory that people can hear better when they have just eaten a large meal.
Question 7 options:
A. support 

B. do not support 
1)
B. greater than
2)
B. the same as or less than
3)
tTest: TwoSample Assuming Equal Variances  
big  small  
Mean  23.33333333  21.33333333 
Variance  4.333333333  4.333333333 
Observations  3  3 
Pooled Variance  4.333333333  
Hypothesized Mean Difference  0  
df  4  
t Stat  1.176696811  
P(T<=t) onetail  0.152279392  
t Critical onetail  2.131846786  
P(T<=t) twotail  0.304558785  
t Critical twotail  2.776445105 
t = 1.1767 (t Stat)
4)
critical value = 2.1318 { one tail)
5)
since TS < critical value
we don't reject the null
6)
insufficient
7)
do not support
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