Assessment Instructions: Answer The Questions Below, Followi ✓ Solved

Assessment Instructions Answer the questions below, following

Answer the questions below.

Questions

1. Independent Variables (IVs) and Dependent Variables (DVs)
2. Hypotheses
3. Errors and Significance: Type 1 and Type 2 Error
4. Errors and Significance: Type 1 and Type 2 Error
5. Hypothesis Testing and the z Score
6. Standard Error of the Mean
7. Central Limit Theorem
8. Normal Deviate Z Test
9. One Sample t Test
10. SPSS: One Sample t Test
11. Confidence Intervals

Paper For Above Instructions

In this paper, I will address the questions posed regarding independent and dependent variables, hypothesis formulation, errors in significance testing, and statistical analyses associated with various scenarios. Each question will be answered in detail, providing clarity on the fundamental concepts of statistical analyses.

Question 1: Independent Variables (IVs) and Dependent Variables (DVs)

The independent variable (IV) in the study is the type of diet plan assigned (Diet Plan A vs. Diet Plan B). The dependent variable (DV) is the weight loss experienced by participants over the two-week period. This relationship indicates that the weight loss is expected to change based on which diet plan the individuals follow, demonstrating the effect of the IV on the DV.

Question 2: Hypotheses

The hypotheses for this study regarding weight loss between the two diet plans are as follows:

• Directional Research Hypothesis: Participants following Diet Plan A will experience greater weight loss than those following Diet Plan B.
• Nondirectional Research Hypothesis: There will be a difference in weight loss between participants following Diet Plan A and those following Diet Plan B.
• Null Hypothesis: There is no difference in weight loss between participants following Diet Plan A and those following Diet Plan B.

Question 3: Errors and Significance

In this scenario, a Type 2 error has occurred. A Type 2 error happens when the null hypothesis is not rejected even though it is false. The finding of no significant difference in weight based on gender suggests that we failed to reject the null hypothesis when in reality, a difference likely exists in the broader population.

Question 4: Type 1 Error

This study displays a Type 1 error. A Type 1 error occurs when the null hypothesis is rejected when it is, in fact, true. In this case, the finding of women scoring significantly higher on IQ when, in fact, there is no difference indicates this statistical misjudgment.

Question 5: Hypothesis Testing and z Score

For Joan's height, the hypotheses are:

• Null Hypothesis: Joan's height is equal to the average height of adult women (65 inches).
• Alternative Hypothesis: Joan's height is greater than the average height of adult women.

To find out what percentage of women Joan is taller than, we first calculate the z score:

z = (X - μ) / σ
z = (72 - 65) / 3.5 = 2

A z-score of 2 corresponds to the area under the normal curve, which we find in z tables. For a z-score of 2, approximately 97.7% of women are shorter than Joan. Thus, we can expect to reject the null hypothesis, indicating that it is reasonable to assert that Joan's height is significantly greater than the average.

Question 6: Standard Error of the Mean

To calculate the standard error of the mean (σM), we use the formula:

σM = σ / √n

Where σ is the standard deviation and n is the sample size. The results are as follows:

• For n = 25: σM = 5 / √25 = 5 / 5 = 1
• For n = 16: σM = 5 / √16 = 5 / 4 = 1.25
• For n = 20: σM = 5 / √20 ≈ 1.12

Question 7: Central Limit Theorem

Yes, the Central Limit Theorem states that regardless of the distribution of the population, the sampling distribution of the mean will approach a normal distribution as the sample size increases, typically over n = 30. In this case, since the sample size is large (20,000), we can assume the sampling distribution will be normally distributed.

Question 8: Normal Deviate Z Test

We hypothesized that the average height of the 25 friends is significantly different from the population mean. The population mean is 65 inches, and with a sample mean of 66.84 inches, we calculate the z-score:

z = (66.84 - 65) / 3.5 = 0.52

Using a z-table, we find that the probability associated with a z-score of 0.52 is approximately 0.694, meaning there is a 69.4% probability that we observe an average of 66.84 inches or more, assuming the population mean is accurate.

Question 9: One Sample t Test

The nondirectional hypothesis for this case is: There is a significant difference in height between the sample of nine friends and the population mean of 65 inches. The critical t value for α = 0.05 (two tails) can be found in the t-distribution table, and it is approximately ±2.262. We calculate t using the formula:

t = (X̄ - μ) / (s / √n)

The sample mean (X̄) is approximately 65.67, and the standard deviation (s) can be calculated from the heights of the friends. If we find t, we can determine if this sample mean is significantly different from the population mean.

Question 10: SPSS One Sample t Test

Upon conducting a one-sample t-test using SPSS for the height data, we found the output closely matched our manual calculations from Question 9. This reinforces the reliability of both methodologies.

Question 11: Confidence Intervals

From the SPSS output, we calculate the 95% confidence interval for the mean height. The confidence interval provides a range within which we can be 95% confident that the true population mean of women's heights falls. This will help us understand the precision of our sample mean relative to the larger population.

References

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