Below are six data sets with 6 randomly selected scores in each data set. Your task is to determine if the scores were drawn from a population with μ = 5. Before you calculate the one-sample t test for each sample, make a guess as to whether or not you think the sample came from the population with μ = 5. After you make your guess, calculate a one-sample t test with α = .05 and a two-tailed test of significance. Then, based on t obs, indicate the decisions you would make for the statistical hypotheses.



Set A

Set B

Set C

Set D

Set E

Set F

5

6

8

5

4.75

5

6

3

9

5

3.25

6

7

3

1

5

3.00

7

5

4

8

4

3.25

8

6

3

9

6

4.75

7

7

5

7

5

5.00

6

To determine if the scores in each data set were drawn from a population with μ = 5, we can perform a one-sample t-test.

Before calculating the t-test, let's make a guess as to whether or not each sample came from the population with μ = 5.

Set A: Based on the scores (5, 6, 8, 5, 4.75, 5), a guess could be made that this sample was drawn from a population with μ = 5.

Now, let's calculate the one-sample t-test for Set A.

Step 1: Calculate the sample mean (x̄):
x̄ = (5 + 6 + 8 + 5 + 4.75 + 5)/6 = 5.4583

Step 2: Calculate the sample standard deviation (s):
s = √[((5 - 5.4583)^2 + (6 - 5.4583)^2 + (8 - 5.4583)^2 + (5 - 5.4583)^2 + (4.75 - 5.4583)^2 + (5 - 5.4583)^2)/(6-1)] = 1.0625

Step 3: Calculate the t-value:
t = (x̄ - μ)/(s/√n) = (5.4583 - 5)/(1.0625/√6) = 0.3992

Step 4: Determine the degrees of freedom (df):
df = n - 1 = 6 - 1 = 5

Step 5: Look up the critical t-value at α = .05 for a two-tailed test and df = 5. From the t-distribution table, the critical t-value is approximately ±2.571.

Step 6: Make a decision:
Since the calculated t-value (0.3992) is within the range of -2.571 and 2.571, we fail to reject the null hypothesis. There is not enough evidence to suggest that this sample is significantly different from a population with μ = 5.

Repeat these steps for Sets B, C, D, E, and F.

Set B:
Guess: Sample was not drawn from a population with μ = 5.
x̄ = 4.6667, s = 1.7529, t = -2.6753, df = 5, Decision: Reject the null hypothesis.

Set C:
Guess: Sample was not drawn from a population with μ = 5.
x̄ = 4.9167, s = 1.8257, t = -1.1841, df = 5, Decision: Fail to reject the null hypothesis.

Set D:
Guess: Sample was not drawn from a population with μ = 5.
x̄ = 4.6667, s = 2.4833, t = -1.1037, df = 5, Decision: Fail to reject the null hypothesis.

Set E:
Guess: Sample was not drawn from a population with μ = 5.
x̄ = 6, s = 2.5166, t = 1.9948, df = 5, Decision: Fail to reject the null hypothesis.

Set F:
Guess: Sample was drawn from a population with μ = 5.
x̄ = 5, s = 0.9984, t = 0, df = 5, Decision: Fail to reject the null hypothesis.

In summary, based on the calculated t-values, we would reject the null hypothesis for Set B and fail to reject the null hypothesis for Sets A, C, D, E, and F.

To determine if the scores in each data set were drawn from a population with a mean of 5, we need to conduct a one-sample t-test.

Before calculating the t-test, let's make a guess. Based on the provided data, it's difficult to make a precise guess without conducting the statistical analysis. However, we can make an initial assumption that the samples might come from a population with a mean of 5 since the mean of the sample is close to 5 in most sets.

To perform the one-sample t-test, we will calculate the t-statistic using the following formula:

t = (sample mean - population mean) / (standard deviation / √sample size)

Now let's calculate the t-test for each data set using α = 0.05 (significance level) and a two-tailed test of significance.

Set A:
Sample mean = 5
Sample standard deviation = √[(5-5)^2 + (6-5)^2 + (8-5)^2 + (5-5)^2 + (4.75-5)^2 + (5-5)^2] / (6-1) ≈ 1.01
Sample size = 6

t = (5 - 5) / (1.01 / √6) ≈ 0

Set B:
Sample mean = 6
Sample standard deviation = √[(6-5)^2 + (3-5)^2 + (9-5)^2 + (5-5)^2 + (3.25-5)^2 + (6-5)^2] / (6-1) ≈ 2.23
Sample size = 6

t = (6 - 5) / (2.23 / √6) ≈ 0.69

Set C:
Sample mean = 3
Sample standard deviation = √[(3-5)^2 + (1-5)^2 + (5-5)^2 + (3-5)^2 + (7-5)^2 + (5-5)^2] / (6-1) ≈ 2.11
Sample size = 6

t = (3 - 5) / (2.11 / √6) ≈ -1.42

Set D:
Sample mean = 4.67
Sample standard deviation = √[(4.67-5)^2 + (8-5)^2 + (4-5)^2 + (3.25-5)^2 + (8-5)^2 + (6-5)^2] / (6-1) ≈ 1.76
Sample size = 6

t = (4.67 - 5) / (1.76 / √6) ≈ -0.44

Set E:
Sample mean = 4.42
Sample standard deviation = √[(4.42-5)^2 + (7-5)^2 + (7-5)^2 + (5-5)^2 + (7-5)^2 + (5-5)^2] / (6-1) ≈ 1.34
Sample size = 6

t = (4.42 - 5) / (1.34 / √6) ≈ -1.79

Set F:
Sample mean = 5.75
Sample standard deviation = √[(5.75-5)^2 + (7-5)^2 + (4-5)^2 + (3.25-5)^2 + (7-5)^2 + (6-5)^2] / (6-1) ≈ 1.71
Sample size = 6

t = (5.75 - 5) / (1.71 / √6) ≈ 0.83

Based on the calculated t-values for each data set, we can make decisions for the statistical hypotheses:

- Set A: Since t ≈ 0 and it is not significant, we fail to reject the null hypothesis.
- Set B: Since t ≈ 0.69 and it is not significant, we fail to reject the null hypothesis.
- Set C: Since t ≈ -1.42 and it is significant, we reject the null hypothesis.
- Set D: Since t ≈ -0.44 and it is not significant, we fail to reject the null hypothesis.
- Set E: Since t ≈ -1.79 and it is significant, we reject the null hypothesis.
- Set F: Since t ≈ 0.83 and it is not significant, we fail to reject the null hypothesis.

In conclusion, based on the t-values, we reject the null hypothesis for Sets C and E, indicating that they likely do not come from a population with a mean of 5. For the other sets, we fail to reject the null hypothesis, suggesting they might come from a population with a mean of 5.