You ask 16 people in your Statistics class what their grade is. The data appears to be distributed normally. You find a sample mean and sample standard deviation of 60 and 24, respectively. Construct and interpret a 95% confidence interval for the population mean class grade.
Table of contents
- 1. Intro to Stats and Collecting Data55m
- 2. Describing Data with Tables and Graphs1h 55m
- 3. Describing Data Numerically1h 45m
- 4. Probability2h 16m
- 5. Binomial Distribution & Discrete Random Variables2h 33m
- 6. Normal Distribution and Continuous Random Variables1h 38m
- 7. Sampling Distributions & Confidence Intervals: Mean1h 53m
- 8. Sampling Distributions & Confidence Intervals: Proportion1h 12m
- 9. Hypothesis Testing for One Sample2h 19m
- 10. Hypothesis Testing for Two Samples3h 22m
- 11. Correlation1h 6m
- 12. Regression1h 4m
- 13. Chi-Square Tests & Goodness of Fit1h 20m
- 14. ANOVA1h 0m
7. Sampling Distributions & Confidence Intervals: Mean
Confidence Intervals for Population Mean
Problem 7.2.34b
Textbook Question
Finite Population Correction Factor If a simple random sample of size n is selected without replacement from a finite population of size (n>0.05N), and the sample size is more than 5% of the population size , better results can be obtained by using the finite population correction factor, which involves multiplying the margin of error E by [Image]. Refer to the weights of the M&M candies in Data Set 38 “Candies” in Appendix B.
b. Use only the red M&Ms and treat that sample as a simple random sample selected from the population of the 345 M&Ms listed in the data set. Find the 95% confidence interval estimate of the mean weight of all 345 M&Ms. Compare the result to the actual mean of the population of all 345 M&Ms.

1
Step 1: Identify the given values and conditions. From the problem, the population size (N) is 345, and the sample size (n) is such that n > 0.05N, meaning the sample size is more than 5% of the population. This condition necessitates the use of the finite population correction factor.
Step 2: Recall the formula for the finite population correction factor (FPCF), which is: \( \sqrt{\frac{N - n}{N - 1}} \). Here, N is the population size, and n is the sample size. This factor will be used to adjust the margin of error.
Step 3: Calculate the margin of error (E) for the confidence interval. The formula for E is: \( E = z \cdot \frac{s}{\sqrt{n}} \), where z is the z-score corresponding to the 95% confidence level (typically 1.96), s is the sample standard deviation, and n is the sample size. Compute this value first without applying the correction factor.
Step 4: Adjust the margin of error using the finite population correction factor. Multiply the previously calculated margin of error (E) by the correction factor \( \sqrt{\frac{N - n}{N - 1}} \). This gives the corrected margin of error.
Step 5: Construct the 95% confidence interval for the mean weight of the M&Ms. The formula is: \( \text{Confidence Interval} = \bar{x} \pm E \), where \( \bar{x} \) is the sample mean and E is the corrected margin of error. Compare the resulting confidence interval to the actual mean of the population to evaluate the accuracy of the estimate.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Finite Population Correction Factor
The Finite Population Correction Factor (FPC) is used when sampling without replacement from a finite population. It adjusts the margin of error to account for the reduced variability in the sample when a significant portion of the population is sampled. Specifically, when the sample size exceeds 5% of the population, the FPC helps provide a more accurate estimate of the population parameters.
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Confidence Interval
A confidence interval is a range of values, derived from sample statistics, that is likely to contain the true population parameter with a specified level of confidence, typically 95%. It is calculated using the sample mean, the standard error, and a critical value from the normal distribution. This interval provides insight into the precision of the sample estimate and the uncertainty associated with it.
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Simple Random Sampling
Simple random sampling is a fundamental sampling technique where each member of the population has an equal chance of being selected. This method ensures that the sample is representative of the population, minimizing bias. In the context of the question, treating the selected red M&Ms as a simple random sample allows for valid statistical inferences about the mean weight of all M&Ms in the population.
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