Posted by **Lily** on Thursday, February 28, 2008 at 1:37am.

Can someone please explain to me the procedures for getting an answer to the questions. Step by step would be great since I'm very confused. I know it's probably simple.

* Numerous studies have demonstrated that listening to music while studying can improve memory. To demonstrate this phenomena, a researcher obtains a sample of college students and gives them a standardized memory test while they are listening to background music. under normal circumstances(without music), the scores on the test form a normal-shaped distribution with a mean of lower case mu 25 and a standard deviation of lower case sigma 6. The sample produces a mean score of M=28.

a)If the sample consists of n=4 students, is this result sufficient to conclude that music had a significant effect on memory scores? Use a two-tailed test with alpha level=.05

b) If the sample consists of n=36 students, is this result sufficient to conclude that the music had a significant event? Again, use a two test with alpha level= .05.

- statistics -
**MathGuru**, Thursday, February 28, 2008 at 9:58am
Both parts are using one-sample two-tailed tests. Since the sample in part a) is so small, you can try a one-sample t-test and for part b), a one-sample z-test.

For the t-test:

t-statistic = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)

Ho: µ = 25 -->null hypothesis

Ha: µ does not equal 25 -->alternate hypothesis

You can finish this calculation:

t-statistic = (28 - 25)/(6/√4)

Using a t-table at 0.05 level of significance for a two-tailed test (alternate hypothesis does not show a specific direction) at 3 degrees of freedom (df = n - 1 = 4 - 1 = 3), look for the value and remember it could be + or - that value because the results could be in either tail of the distribution.

Does the test statistic exceed the negative or positive critical or cutoff value from the t-table? If it does, the null is rejected in favor of the alternate hypothesis and you can conclude a significant difference. If it does not, then you cannot reject the null and conclude a difference.

For the z-test, everything is basically the same (even the formula), except for the sample size and using the z-table instead of a t-table. Here's a hint: for a two-tailed test at alpha level .05 , the cutoff values are +/-1.96 using a z-table.

I hope this will help get you started.

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