Question

show all work please

A professor predicts that students will lean better if they study to white noise (similar to a fan) compared to music or no sound. She randomly divides 27 students into three groups and sends them to three different rooms. In the first room, students hear white noise, in the second, music from a local radio station, and in the third, the door is closed to help block out normal sound from the hall. Each group is given 30 minutes to study a section of their text after which they take a 10-question multiple- choice exam. Table 31.5 contains the results. 1. White Noise No Sound Music 6 2 5 10 Table 31.5. Test results. a. Calculate the mean test score for each group. Calculate the standard deviation of the test scores for each group. Open a new data table in SPSS - Make columns 1, 2 and 3 to be White Noise, Music and No Sound respectively Select ANALYZE> DESCRIPTIVE STATISTICS> DESCRIPTIVESS Select the variables to analyze and click OK b. Make comparative boxplots for the test results of the three groups. Do you think that the boxplots give sufficient evidence that there is a difference in population mean test results depending on the type of noise? Explain. (Hint: In SPSS, select GRAPHS LEGACY DIALOGS>BOXPLOT> SUMMARY OF SEPARATE VARIABLES> DEFINE) c. Run an ANOVA. State the hypotheses you are testing. Show the calculations for the F-statistic. What are the degrees of freedom associated with this F-statistic? (Hint: In SPSS, restructure the data similar to the class activity assignment) - In a new column (4th column), paste (or type) the values of white noise, music and no sound in that order. - In another column (5th column), type the number one from row 1 to 9, the number 2 in rows 10 to 18, the number 3 in rows 19 to 27. The numbers on the fifth column represent the treatments, 1: White noise, 2: Music and 3: No Sound. You can add labels in the VARIABLE VIEW tab. - Select ANALYZE>COMPARE MEANS ONE-WAY ANOVA Send the fourth column to the DEPENDENT LIST and the fifth column to FACTOR Select OK. Find the F-statistic and degrees of freedom in the output ANOVA table d. What is the p-value of the test statistic? What is your condlusion? (Hint: Use the ANOVA table from (c)

Answer 1

data

white	Music	No sound
8	5	4
5	7	7
5	3	6
7	5	2
8	5	4
7	8	3
8	4	5
3	5	5
10	7	4

a)

Groups	Average	sd
white	6.7778	2.1082
Music	5.4444	1.5899
No sound	4.4444	1.5092

b)

3 2

yes there is sufficient evidence for difference

c)

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
white	9	61	6.777777778	4.444444
Music	9	49	5.444444444	2.527778
No sound	9	40	4.444444444	2.277778
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	24.66666667	2	12.33333333	4	0.031676	3.402826
Within Groups	74	24	3.083333333
Total	98.66666667	26

F = 4

d)

p-value = 0.031676

p-value < alpha (0.05)

hence we reject the null hypothesis