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Difference between two means (Independent Samples): hypothesis tests and confidence intervals.
- Do piano lessons improve the spatial-temporal reasoning of preschool children? Below is the score on a spatial reasoning test of 34 preschool children after six months of piano lessons.
2 5 7 -2 2 7 4 1 0 7 3 4 3 4 9 4 5 0 3 6 -1 3 4 6 7 -2 7 -3 3 4 4 2 9 6
Below is the score made on the same test by 44 preschool children in a control group. 1 -1 0 1 -4 0 0 1 0 -1 0 1 1 -3 -2 4 -1 2 4 2 2 2 -3 -3 0 2 0 -1 3 -1 5 -1 7 0 4 0 2 1 -6 0 2 -1 0 -
(Data from F. H. Rauscher et al., āMusic training causes long-term enhancement of preschool childrenās spatial-temporal reasoning,ā Neurological Research, 19 (1997), pp. 2-8.)
a. Display the data with histograms and boxplots and summarize the distributions in a paragraph. You are interested in:
- Does the sample of each seem to be a normal distribution?
ANSWER: The histogram of the piano lesson group is symmetric and unimodal. The histogram of the control group is unimodal and only slightly skewed (changing XSCL to exactly 2 reveals a symmetric shape).
- Does either data set have outliers?
ANSWER: The boxplot of the piano lesson group shows no outliers. The boxplot of the control group reveals two outliers, one at each end.
b. Define Population 1: The spatial-temporal reasoning scores of all preschool children that have taken six months of piano lessons. Population 2: The spatial-temporal reasoning scores of all preschool children that have NOT taken six months of piano lessons.
Ī¼ 1 : The average spatial-temporal reasoning score of all preschool children
that have taken six months of piano lessons.
Ī¼ 2 : The average spatial-temporal reasoning score of all preschool children
that have NOT taken six months of piano lessons.
c. Make a table with the sample size, sample mean, and sample standard deviation for each of the two groups.
ANSWER:
Sample Size Sample Mean Sample Standard Deviation Population 1 (^) n 1 (^) = 34 x 1 (^) = 3. 617 S 1 = 3. 0552 Population 2 (^) n 2 (^) = 44 x (^) 2 = 0. 386 S 2 = 2. 4229
d. Are all of the assumptions satisfied so that we may use two-sample t methods?
ANSWER: Yes. We will assume that both of the samples were drawn randomly. Since n 1 (^) = 34 > 15 and the first sample has a normal shape (according to its histogram) and no outliers (according to its boxplot) then the conditions for the first sample are met. Even though there might be some slight skewness in the second sample (according to its histogram) and there are two outliers (according to its boxplot), having n 2 (^) = 44 > 40 shows us that the conditions for the second sample are met. In conclusion, we note that the two populations (described in part b) are independent of each other. Now all of the assumptions are satisfied so that we may use two-sample t methods.
e. Write out the six steps in a test of hypothesis to prove that piano lessons improve spatial reasoning in preschool children. Test at the 5% level.
ANSWER:
1 2
0 1 2 :
Ī¼ Ī¼
Ī¼ Ī¼
H a
H
Ī±= 0. 05
Sample Size Sample Mean Sample Standard Deviation Population 1 (^) n 1 (^) = 34 x 1 (^) = 3. 617 S 1 = 3. 0552 Population 2 (^) n 2 (^) = 44 x (^) 2 = 0. 386 S 2 = 2. 4229
- The Chapin Social Insight Test is a psychological test designed to measure how accurately the subject appraises other people. The possible scores on the test range from 0 to 41. During the development of the Chapin test, it was given to several different groups of people. Here are the results for male and female college students majoring in the liberal arts:
Group Sex n^ x (^) Sx 1 Male 133 25.34 5. 2 Female 162 24.94 5.
Do these data support the contention that female and male students differ in average social insight? Test at the 5% level.
ANSWER:
1 2
0 1 2 :
Ī¼ Ī¼
Ī¼ Ī¼ ā
H a
H
- Ī±= 0. 05
- See table above.
- Test statistic is t = 0. 654
P-value = 0.
5. P-value Ī±
0.514 > 0. do NOT reject H 0
- There is not significant evidence of a gender difference in mean social insight score.
- Music and memory. Is it a good idea to listen to music when studying for a big test? In a study conducted by some Statistics students, 62 people were randomly assigned to listen to rap music, music by Mozart, or no music while attempting to memorize objects pictured on a page. They were then asked to list all the objects they could remember. Here are summary statistics for each group: Rap Mozart No Music Count^29 20 Mean 10.72^ 10.0^ 12. StDev 3.99^ 3.19^ 4.
Create a 90% confidence interval for the mean difference in memory score between students who study to Mozart and those who listen to no music at all. Interpret your interval.
ANSWER: Let the Mozart scores be population 1 and no music be population 2. The confidence interval is (-5.351 , -0.189).
The confidence interval takes x 1 ā x 2 and uses it to estimate Ī¼ 1 ā Ī¼ 2. Soā¦
We are 90% confident that the difference between the mean number of objects remembered by those who listen to Mozart and the mean number of objects of those who listen to no music is between 0.189 and 5.352 objects. In other words, we are 90% confident that the mean number of objects remembered by those who listen to Mozart is between 0.189 and 5.352 objects lower than the mean of those who listened to no music.
