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)
t-Test: Two-Sample 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) one-tail | 0.152279392 | |
t Critical one-tail | 2.131846786 | |
P(T<=t) two-tail | 0.304558785 | |
t Critical two-tail | 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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