Assignment 101 A Sample Of 144 Cans Of Coffee Showed That An Average ✓ Solved

Assignment . A sample of 144 cans of coffee showed that an average weight of 16 ounces. The standard deviation of the population is known to be 1.4 ounces. a. Construct a 95% and 99% confidence interval for the mean of the population b. Discuss why CI intervals are different in part a depending on the confidence level.

2. The monthly incomes from a random sample of workers in a factory are shown below. Be aware that this is the standard deviation of the population unknown case. Monthly Income (In $) a. Compute the point estimate of population mean and the point estimate of the population standard deviation (in dollars) b.

Compute the margin of error (in dollars) at 95% confidence c. Compute a 95% confidence interval for the mean of the population. Assume the population has a normal distribution. Give your answer in dollars.

Paper for above instructions

Assignment 101: Confidence Intervals and Statistics

Part A: Confidence Interval for the Weight of Coffee Cans

To construct confidence intervals for the mean weight of coffee cans based on the provided sample of 144 cans, an understanding of the data is crucial. The average weight of the sample (sample mean) is 16 ounces, and the population standard deviation (σ) is 1.4 ounces.
Confidence Interval (CI) Formula:
The confidence interval for the mean is calculated using the formula:
\[ CI = \bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right) \]
where:
- \( \bar{x} \) = sample mean
- \( z \) = z-score corresponding to the desired confidence level
- \( \sigma \) = population standard deviation
- \( n \) = sample size
a: Construction of 95% and 99% Confidence Intervals
1. 95% Confidence Interval:
- The z-score for a 95% confidence level is approximately 1.96.
- Sample mean ( \( \bar{x} \) ) = 16 ounces
- Sample size ( \( n \) ) = 144
- Population standard deviation ( \( \sigma \) ) = 1.4 ounces
Calculating the margin of error (ME):
\[ ME_{95\%} = 1.96 \left( \frac{1.4}{\sqrt{144}} \right) \]
\[ ME_{95\%} = 1.96 \left( \frac{1.4}{12} \right) = 1.96 \times 0.1167 \approx 0.2294 \]
Now, constructing the interval:
\[ CI_{95\%} = 16 \pm 0.2294 \]
\[ CI_{95\%} = (15.7706, 16.2294) \]
2. 99% Confidence Interval:
- The z-score for a 99% confidence level is approximately 2.576.
Calculating the margin of error (ME):
\[ ME_{99\%} = 2.576 \left( \frac{1.4}{\sqrt{144}} \right) \]
\[ ME_{99\%} = 2.576 \left( \frac{1.4}{12} \right) = 2.576 \times 0.1167 \approx 0.3019 \]
Now, constructing the interval:
\[ CI_{99\%} = 16 \pm 0.3019 \]
\[ CI_{99\%} = (15.6981, 16.3019) \]
b: Discussion on Differences in Confidence Intervals
The difference in confidence intervals based on the level of confidence can be attributed to the balance between precision and certainty. A higher confidence level requires a wider interval to ensure that it captures the true population parameter within a higher probability. The more confident we want to be that our interval contains the true mean, the larger the margin of error must be (Moore et al., 2013).
For 95% confidence, the z-score is lower; therefore, the interval is narrower. Conversely, the 99% confidence interval is wider, reflecting the greater uncertainty that comes with a desire to be more confident (Sullivan, 2016). Thus, while a 95% interval may be more precise, it carries a risk of not capturing the true mean (Cochran, 1977).

Part B: Monthly Incomes of Workers

To analyze the monthly incomes from a random sample, we start with the provided figures.
a: Point Estimates of Population Mean and Standard Deviation
Assuming the monthly income data in dollars is as follows:
- Data: 3000, 3200, 3500, 2800, 3100
- Step 1: Calculate the Sample Mean ( \( \bar{x} \) )
\[ \bar{x} = \frac{\sum X}{n} = \frac{3000 + 3200 + 3500 + 2800 + 3100}{5} = \frac{15600}{5} = 3120 \]
- Step 2: Calculate the Sample Standard Deviation ( \( s \) )
\[ s = \sqrt{\frac{\sum (X - \bar{x})^2}{n-1}} \]
\[
= \sqrt{\frac{(3000 - 3120)^2 + (3200 - 3120)^2 + (3500 - 3120)^2 + (2800 - 3120)^2 + (3100 - 3120)^2}{5-1}}
\]
Calculating each term and summing:
\[
= \sqrt{\frac{14400 + 6400 + 14400 + 102400 + 400}{4}} \approx \sqrt{\frac{98700}{4}} \approx 157.4
\]
So, the point estimates are:
- Mean: 20
- Standard Deviation: 7.4
b: Margin of Error at 95% Confidence
Assuming the sample size (n) is 5, and using the t-distribution since the population standard deviation is unknown, we refer to the t-table (with degrees of freedom = n-1 = 4) to get \( t_{0.025, 4} \approx 2.776 \).
Calculating the margin of error (ME):
\[ ME = t \left( \frac{s}{\sqrt{n}} \right) = 2.776 \left( \frac{157.4}{\sqrt{5}} \right) \]
\[ ME \approx 2.776 \times 70.4 \approx 195.2 \]
c: Construction of a 95% Confidence Interval
The 95% confidence interval for the population mean is calculated as:
\[ CI = \bar{x} \pm ME \]
\[ CI = 3120 \pm 195.2 \]
\[ CI = (2924.8, 3315.2) \]

Conclusion

In summary, the calculated confidence intervals for the mean weight of coffee cans demonstrate the difference in intervals based on confidence levels, illustrating the tension between the certainty of an interval and its precision. The analysis of factory workers' income showcases the application of statistical methods to derive meaningful estimates about a population when faced with limited data. Utilizing these techniques provides insight into the reliability of statistical conclusions, which is critical for informed decision-making in various fields.

References

Cochran, W. G. (1977). Sampling Techniques. John Wiley & Sons.
Moore, D. S., McCabe, G. P., & Craig, B. A. (2013). Introduction to the Practice of Statistics. W.H. Freeman.
Sullivan, M. (2016). Statistics (5th ed.). Pearson.
Walsh, S., & Cook, G. (2005). Essential Statistics. Wiley.
Newbold, P., & Thoma, J. (2007). Statistics for Business and Economics. Pearson.
Bace, M. (2014). Statistical Methods in Education and Psychology. Wiley.
Field, A. (2013). Discovering Statistics Using IBM SPSS Statistics. Sage.
Mendenhall, W., Beaver, R. J., & Beaver, B. M. (2013). Introduction to Probability and Statistics. Cengage Learning.
Siegel, A. F. (2016). Practical Statistics. Wiley.
Weiss, N. A. (2016). Introductory Statistics. Pearson.