A Quality Control Department Took A Sample Of 40 Products

1 A Quality Control Department Noted Took A Sample Of 40 Product

A quality control department took a sample of 40 products and found the mean number of defects to be 6.15, with a standard deviation of 2.53. The distribution was known to be normal. Construct a 95 percent confidence interval for the mean number of defects.

A corporation is trying to estimate the mean amount an employee spends on dental care. They surveyed 45 employees, and found a mean amount spent of $1820, with a sample standard deviation of $660. Find a 95% confidence interval for the mean amount spent.

Pepsi wants to estimate the average number of cans of Pepsi consumed. The sample consists of 16 people, with a mean of 60 cans and a standard deviation of 20 cans.

  • a. Is the t or z distribution used here? Why?
  • b. What is the value of the test statistic to be used for a 90% confidence interval?
  • c. Create a 90% confidence interval.
  • d. Could Pepsi be reasonable in concluding that the population mean is 59 cans?

Last year, a survey showed that 30% of travelers stay in budget hotels. The hotel association wants to update this proportion.

  • a. The new study is to use the 90 percent confidence level. The estimate is to be within 1 percent of the population proportion. What is the necessary sample size?
  • b. If we wanted to reduce the required sample size, what could we do?

A survey of singles was made regarding their restaurant eating habits. A sample of 60 singles had a mean number of restaurant visits of 2.76 meals per week. The standard deviation of the sample was 0.75 meals per week. Create a 98 percent confidence interval for the mean of the population.

Paper For Above instruction

Understanding and accurately estimating population parameters through confidence intervals is a fundamental aspect of statistical inference. This paper addresses multiple scenarios involving confidence intervals, hypothesis testing, and sample size determinations across different contexts, illustrating key concepts with detailed calculations and interpretations.

1. Confidence Interval for the Mean Number of Defects in Products

The quality control department’s sample of 40 products yielded a sample mean of 6.15 defects, with a standard deviation of 2.53, based on a normal distribution. To estimate the true population mean with a 95% confidence level, we employ the t-distribution because the sample size is less than 30, or if the population standard deviation is unknown, which is often the case in real-world scenarios. Given the sample size of 40 (which is large), using the z-distribution is also acceptable, but standard practice favors the t-distribution here.

The critical value for a 95% confidence interval with a t-distribution and 39 degrees of freedom is approximately 2.022. The standard error (SE) is calculated as the sample standard deviation divided by the square root of the sample size: SE = 2.53 / √40 ≈ 0.4. The confidence interval (CI) is then given by:

CI = mean ± (critical value × SE) = 6.15 ± (2.022 × 0.4) ≈ 6.15 ± 0.809

Therefore, the 95% CI for the average number of defects is approximately (5.341, 6.959).

2. Confidence Interval for Employee Dental Care Expenses

A survey of 45 employees revealed a mean annual expenditure of $1820 with a sample standard deviation of $660. Since the population standard deviation is unknown and the sample size exceeds 30, the t-distribution is appropriate. The degrees of freedom are 44. The critical t-value for a 95% confidence level is approximately 2.015.

Calculating the standard error: SE = 660 / √45 ≈ 98.54. The confidence interval is:

CI = 1820 ± (2.015 × 98.54) ≈ 1820 ± 198.6

Thus, the 95% CI ranges approximately from $1621.4 to $2018.6, suggesting with high confidence that the true mean expenditure lies within this interval.

3. Estimating Average Number of Cans of Pepsi Consumed

The sample of 16 individuals indicates a mean of 60 cans, with a standard deviation of 20 cans. Since the sample size is small (

  • a. The t-distribution is used because the sample size is less than 30 and the population standard deviation is unknown, making the t-test appropriate.
  • b. For a 90% confidence interval with 15 degrees of freedom, the critical t-value is approximately 1.753.
  • c. Calculating the standard error: SE = 20 / √16 = 20 / 4 = 5. The confidence interval is:

CI = 60 ± 1.753 × 5 = 60 ± 8.765, which yields approximately (51.235, 68.765).

  • d. To determine whether it is reasonable to conclude the population mean is 59 cans, we examine if 59 lies within the confidence interval. Since (51.235, 68.765) contains 59, it is entirely plausible that the true mean equals 59 cans.

4. Updating the Proportion of Travelers Staying in Budget Hotels

Last year, 30% of travelers stayed in budget hotels. The hotel association wishes to estimate this proportion with 90% confidence and within 1% margin of error (0.01).

  • a. The sample size n required for estimating a proportion p with margin of error E at confidence level 1−α can be calculated using:

n = (Z² × p × (1−p)) / E²

where Z is the critical value for the 90% confidence level (Z ≈ 1.645). Plugging in the values:

n = (1.645² × 0.3 × 0.7) / 0.01² ≈ (2.706 × 0.21) / 0.0001 ≈ 0.568 / 0.0001 = 5680

Thus, approximately 5680 respondents are necessary to estimate the proportion within 1% with 90% confidence.

b. To reduce the required sample size, the hotel association could increase the margin of error, i.e., accept a wider confidence interval, or reduce the confidence level, although the latter diminishes the precision of the estimate.

5. Confidence Interval for the Mean Number of Restaurant Visits

A survey of 60 singles shows a mean of 2.76 restaurant visits per week with a standard deviation of 0.75.

Since the sample size exceeds 30, and the population standard deviation is unknown, the t-distribution applies. For a 98% confidence level, the critical t-value with df = 59 is approximately 2.000.

The standard error: SE = 0.75 / √60 ≈ 0.097.

The confidence interval is:

CI = 2.76 ± 2.000 × 0.097 ≈ 2.76 ± 0.194

Consequently, the 98% confidence interval spans approximately from 2.566 to 2.954 visits per week, indicating high confidence that the true population mean lies within this range.

In conclusion, the above cases demonstrate different applications of confidence intervals, and the importance of selecting the appropriate distribution (t or z) based on sample size and variance knowledge. Properly constructing these intervals enables organizations and researchers to make informed decisions and accurate estimates about the populations they study.

References

  • Agresti, A., & Franklin, C. (2017). Statistics: The Art and Science of Learning from Data. Pearson.
  • Devore, J. L. (2015). Probability and Statistics for Engineering and the Sciences. Cengage Learning.
  • Moore, D. S., McCabe, G. P., & Craig, B. A. (2017). Introduction to the Practice of Statistics. W.H. Freeman.
  • Newbold, P., Carlson, W. L., & Thorne, B. (2013). Statistics for Business and Economics. Pearson.
  • Ott, R. L., & Longnecker, M. (2015). An Introduction to Statistical Methods and Data Analysis. Cengage Learning.
  • Ross, S. M. (2014). Introduction to Probability and Statistics. Academic Press.
  • Snedecor, G. W., & Cochran, W. G. (1989). Statistical Methods. Iowa State University Press.
  • Walpole, R. E., Myers, R. H., Myers, S. L., & Ye, K. (2012). Probability & Statistics for Engineers & Scientists. Pearson.
  • Zar, J. H. (2010). Biostatistical Analysis. Pearson.
  • Kreyszig, E. (2011). Introductory Mathematical Analysis. John Wiley & Sons.

Related Assignments