SOLUTION: Eight college students were randomly divided into 2 groups of 4 each to test whether background music reduces studying capacity. Each person was tasked with memorizing a list of 2
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Question 305683: Eight college students were randomly divided into 2 groups of 4 each to test whether background music reduces studying capacity. Each person was tasked with memorizing a list of 20 words. Group 1 had music playing through earphones they were wearing. Group 2 was not distracted. The following are the number of words correctly remembered by each subject. Test whether the music reduces studying capacity at a .05 level of significance.
Group 1 -- DISTRACTED Group 2 – NOT DISTRACTED
Sample mean = 7
Sample size = 4
Sample std.dev. = 2.5 Sample mean = 14
Sample size = 4
Sample std.dev. = 3.5
3a) Direction = UPPER TAILED / LOWER TAILED / 2-TAILED (circle one)
3b) P-value = ________________
3c) REJECT THE NULL or DO NOT REJECT THE NULL? (Circle one)
3d) Can we conclude that music reduces studying capacity? YES or NO (circle one)
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Eight college students were randomly divided into 2 groups of 4 each to test whether background music reduces studying capacity. Each person was tasked with memorizing a list of 20 words.
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Group 1 had music playing through earphones they were wearing.
Group 2 was not distracted.
-----------------------------------------
The following are the number of words correctly remembered by each subject. Test whether the music reduces studying capacity at a .05 level of significance.
Group 1 -- DISTRACTED Group 2 – NOT DISTRACTED
Sample size = 4
Sample mean = 7
Sample std.dev. = 2.5
-----
Group 2 --- NOT DISTRACTED
Sample size = 4
Sample mean = 14
Sample std.dev. = 3.5
3a) Direction = UPPER TAILED / LOWER TAILED / 2-TAILED (circle one)
"Reduces" is the key word in the problem description:
Ans: left-tail test
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Note:
Ho: u(music)-u(no music) = 0
Ha: u(music)-u(no music) < 0 ("no music" helps)
3b) P-value = ________________
Test Statistic:
t(7-14)= (-7)/sqrt[2.5^2/4 + 3.5^2/4] = -3.2549
p-value = P(t < -3.2549 when df = 5.4) = 0.0100
Note: I used a 2-sample T test on a TI calculator
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3c) REJECT THE NULL or DO NOT REJECT THE NULL? (Circle one)
Since the p-value is less than 5%, reject Ho
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3d) Can we conclude that music reduces studying capacity?
YES or NO (circle one)
I'll leave that to you.
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Cheers,
Stan H.
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