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.

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.

a) To determine if the result is sufficient to conclude that music had a significant effect on memory scores for a sample size of n=4 students, we need to perform a two-tailed test with an alpha level of .05.

Step 1: State the null hypothesis (H0) and alternative hypothesis (H1):
- Null hypothesis (H0): There is no significant effect of music on memory scores.
- Alternative hypothesis (H1): There is a significant effect of music on memory scores.

Step 2: Determine the critical value(s) for the test:
Since it's a two-tailed test with an alpha level of .05, we split the alpha level in half (.025) and look up the corresponding critical value(s) in the t-distribution table for a sample size of n=4 and an alpha level of .025. Using the table, the critical value is approximately ±3.182.

Step 3: Calculate the test statistic:
The test statistic used for this scenario is a t-test. We calculate the t-value by subtracting the population mean (25) from the sample mean (28) and dividing it by the standard deviation of the population (6) divided by the square root of the sample size (n=4).

t = (sample mean - population mean) / (population standard deviation / √sample size)
t = (28 - 25) / (6 / √4)
t = 3 / (6 / 2)
t = 3 / 3
t = 1

Step 4: Compare the test statistic with the critical value:
Since the absolute value of the calculated t-value (1) is less than the critical value (3.182), we fail to reject the null hypothesis.

Step 5: Make a conclusion:
Based on the results, we do not have sufficient evidence to conclude that music had a significant effect on memory scores for a sample size of n=4 students at an alpha level of .05.

b) To determine if the result is sufficient to conclude that music had a significant effect for a sample size of n=36 students, we follow a similar process to the one above.

Step 1: State the null hypothesis (H0) and alternative hypothesis (H1):
- Null hypothesis (H0): There is no significant effect of music on memory scores.
- Alternative hypothesis (H1): There is a significant effect of music on memory scores.

Step 2: Determine the critical value(s) for the test:
Again, since it's a two-tailed test with an alpha level of .05, we split the alpha level in half (.025) and look up the corresponding critical value(s) in the t-distribution table for a sample size of n=36 and an alpha level of .025. Using the table, the critical value is approximately ±2.028.

Step 3: Calculate the test statistic:
Using the same formula as before, we calculate the t-value as follows:
t = (sample mean - population mean) / (population standard deviation / √sample size)
t = (28 - 25) / (6 / √36)
t = 3 / (6 / 6)
t = 3 / 1
t = 3

Step 4: Compare the test statistic with the critical value:
Since the absolute value of the calculated t-value (3) is greater than the critical value (2.028), we reject the null hypothesis.

Step 5: Make a conclusion:
Based on the results, we have sufficient evidence to conclude that music had a significant effect on memory scores for a sample size of n=36 students at an alpha level of .05.

Sure! I can help you with that. Let's go through the steps to find the answers to your questions:

a) To determine if the result is sufficient to conclude that music had a significant effect on memory scores for a sample of n=4 students, we can perform a hypothesis test using a two-tailed test with an alpha level of .05.

Step 1: State the Hypotheses
- Null Hypothesis (H0): The mean memory scores with music (µ_music) are not significantly different from the mean memory scores without music (µ_no_music).
- Alternative Hypothesis (Ha): The mean memory scores with music (µ_music) are significantly different from the mean memory scores without music (µ_no_music).

Step 2: Set the Decision Rule
Since it is a two-tailed test, we will divide the alpha level (.05) by 2 to get the critical value for each tail. In this case, each tail will have an alpha level of .025.

Step 3: Calculate the Test Statistic
To calculate the test statistic, we can use the formula for a Z-test:

Z = (M - µ) / (σ / sqrt(n))

Where:
- M = Sample mean (given as M = 28)
- µ = Population mean without music (given as µ = 25)
- σ = Population standard deviation (given as σ = 6)
- n = Sample size (given as n = 4)

Step 4: Make a Decision
Compare the test statistic to the critical value(s) from Step 2. If the test statistic falls in the rejection region (outside the critical values), we can reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

b) To determine if the result is sufficient to conclude that the music had a significant effect for a sample of n=36 students, we will follow the same steps as above with a larger sample size.

Step 1: State the Hypotheses (same as in part a)
- Null Hypothesis (H0): The mean memory scores with music (µ_music) are not significantly different from the mean memory scores without music (µ_no_music).
- Alternative Hypothesis (Ha): The mean memory scores with music (µ_music) are significantly different from the mean memory scores without music (µ_no_music).

Step 2: Set the Decision Rule (same as in part a)

Step 3: Calculate the Test Statistic
Using the same formula as in part a, calculate the test statistic using the given values:
- M = Sample mean (given as M = 28)
- µ = Population mean without music (given as µ = 25)
- σ = Population standard deviation (given as σ = 6)
- n = Sample size (given as n = 36)

Step 4: Make a Decision (same as in part a)

These steps will help you determine if the obtained results are sufficient to conclude that music had a significant effect on memory scores for the given sample sizes.