Module 4: Content Mastery Testing

Instructions

After studying and reviewing the concepts and practices of Module 4, answer the following questions:

1. A sample size of 35 scores has a mean score of 85 points with a standard deviation of 3.25 points. Is there any statistical evidence that the grade population mean is less than 75 points (α = 5%)? Greater than 85 points (α = 10%)
2. A sample size of 49 students has a mean weight of 205 pounds with a standard deviation of 5.8 pounds. Find a 95% confidence interval for the weight population mean. Is there any statistical evidence that the weight population mean is less than 200 pounds (α = 1%)?
3. The following data represents the annual average salary (\$) of a sample of nine Social Workers.
 42,640 41,830 46,990 50,690 40,950 41,460 48,670 40,090 44,720
• If the data comes from a normally distributed population with µ = \$48,920, Is there any statistical evidence that the average salary of Social Workers is less than \$45 000 for a significance level of 10%?

Be sure to review the academic expectations for your submission.

Submission Instructions:

• Submit your assignment by 11:59 PM Eastern on Sunday.
• Review the rubric to determine how your assignment will be graded.
• Your assignment will be run through Turnitin to check for plagiarism.
• Justify any assumption and show all the steps to solve the exercises.

Module 4: Content Mastery Testing

Introduction

Module 4 delves into the essential concepts and practices of statistical testing, focusing on hypothesis testing and confidence intervals. In this essay, we will address three questions based on the knowledge acquired from this module.

Question 1

Given:

• Sample size (n) = 35
• Sample mean (x̄) = 85 points
• Standard deviation (σ) = 3.25 points
• α = 5% and 10%

Hypothesis Testing:

Null Hypothesis (H₀): Population mean (μ) = 75 points Alternative Hypothesis (H₁): Population mean (μ) ≠ 75 points

Calculations:

Z = (x̄ – μ) / (σ / √n)

For α = 5%: Critical Z-value for a two-tailed test = ±1.96

Z = (85 – 75) / (3.25 / √35) Z ≈ 14.44

Since the calculated Z-value is greater than the critical Z-value, we reject the null hypothesis. There is statistical evidence that the grade population mean is not equal to 75 points.

For α = 10%: Critical Z-value for a two-tailed test = ±1.645

Z = (85 – 75) / (3.25 / √35) Z ≈ 14.44

Since the calculated Z-value is greater than the critical Z-value, we reject the null hypothesis. There is statistical evidence that the grade population mean is not equal to 75 points.

Question 2

Given:

• Sample size (n) = 49
• Sample mean (x̄) = 205 pounds
• Standard deviation (σ) = 5.8 pounds
• Confidence level = 95%
• α = 1%

Confidence Interval:

Formula: Confidence Interval = x̄ ± Z * (σ / √n)

Calculations:

Critical Z-value for a 95% confidence level = ±1.96

Confidence Interval = 205 ± 1.96 * (5.8 / √49) Confidence Interval ≈ 205 ± 1.96 * (5.8 / 7) Confidence Interval ≈ 205 ± 1.48

Thus, the 95% confidence interval for the weight population mean is approximately (203.52, 206.48) pounds.

For α = 1%, the critical Z-value for a one-tailed test = -2.33 (since we are testing if the population mean is less than 200 pounds)

Z = (x̄ – μ) / (σ / √n) Z = (205 – 200) / (5.8 / √49) Z ≈ 4.29

Since the calculated Z-value is greater than the critical Z-value, we reject the null hypothesis. There is statistical evidence that the weight population mean is greater than 200 pounds.

Question 3

Given:

• Sample size (n) = 9
• Data: 42,640, 41,830, 46,990, 50,690, 40,950, 41,460, 48,670, 40,090, 44,720
• Population mean (μ) = \$48,920
• α = 10%

Hypothesis Testing:

Null Hypothesis (H₀): Population mean (μ) = \$45,000 Alternative Hypothesis (H₁): Population mean (μ) < \$45,000

Calculations:

t = (x̄ – μ) / (s / √n)

Where s is the sample standard deviation.

Mean (x̄) = (42,640 + 41,830 + 46,990 + 50,690 + 40,950 + 41,460 + 48,670 + 40,090 + 44,720) / 9 ≈ 44,444.44

Standard Deviation (s) ≈ 3,937.62

t = (44,444.44 – 45,000) / (3,937.62 / √9) t ≈ -0.710

Degrees of freedom (df) = n – 1 = 9 – 1 = 8

Critical t-value for a one-tailed test with df = 8 and α = 10% ≈ -1.860

Since the calculated t-value (-0.710) is greater than the critical t-value (-1.860), we fail to reject the null hypothesis. There is no statistical evidence that the average salary of Social Workers is less than \$45,000 at a significance level of 10%.

Conclusion

Module 4 has equipped us with the tools to conduct hypothesis testing and construct confidence intervals effectively. Through the application of statistical methods, we can draw meaningful conclusions about populations based on sample data, enabling informed decision-making in various fields.