Retail Stores Experience Their Heaviest Returns In December ✓ Solved

Retail stores experience their heaviest returns on December 26

Retail stores experience their heaviest returns on December 26 and December 27 each year. Most are gifts that, for some reason, did not please the recipient. The number of items returned, by a sample of 30 persons at a large discount department store, are observed and the summary statistics below are obtained. n=30 x̄=2.468 s=1.207. Determine a 98% confidence interval for (μ) the population mean number of items returned. Confidence interval = ( , ).

Stated here are some claims or research hypotheses that are to be substantiated by sample data. Identify the null hypothesis (H0) and the alternative hypothesis (H1) in terms of the population mean (μ). The mean time a health insurance company takes to pay claims is less than 20 working days:

Statistics in Retail Returns

Understanding retail returns is crucial for businesses, especially during peak seasons like the holidays. After analyzing the data collected from a sample of 30 persons at a large discount department store, we find the average number of items returned to be 2.468 with a standard deviation of 1.207. To assess the uncertainty around this average, we will construct a 98% confidence interval for the population mean number of items returned.

Calculating the Confidence Interval

The formula for a confidence interval for the population mean (μ) when the population standard deviation is unknown is given by:

CI = x̄ ± t*(s/√n)

In this formula:

  • x̄ = sample mean
  • t* = t-value from the t-distribution table
  • s = sample standard deviation
  • n = sample size

To find the t-value for a 98% confidence interval with n-1 degrees of freedom (29 degrees of freedom), we refer to a t-table. The critical t-value for a two-tailed test at 98% confidence level is approximately 2.462.

Substituting the values into the formula, we get:

CI = 2.468 ± 2.462*(1.207/√30)

Calculating the margin of error:

Margin of error = 2.462 * (1.207/5.477) ≈ 0.5028

So, the confidence interval is:

CI = 2.468 ± 0.5028

This results in approximately:

Lower Limit = 2.468 - 0.5028 ≈ 1.965

Upper Limit = 2.468 + 0.5028 ≈ 2.970

Final Confidence Interval

Thus, the 98% confidence interval for the population mean number of items returned is approximately (1.97, 2.97).

Hypothesis Testing in Claims Processing

Next, we focus on the hypothesis related to the mean time a health insurance company takes to pay claims. This statement can be formulated as:

  • Null Hypothesis (H0): μ ≥ 20 days
  • Alternative Hypothesis (H1): μ

This scenario suggests we are looking to substantiate through data that the average processing time is indeed less than 20 working days. To test this hypothesis, we would conduct a t-test and potentially gather a sample of claims processing times to appraise our null hypothesis accurately.

Assessing Germination Times of Seeds

Let’s now explore the second problem related to the germination times of a new strain of snap beans. Here, we need to calculate a 95% confidence interval for the true mean germination time based on germination time data for seven seeds of the new strain.

Assuming we collected the following germination days: 3, 4, 5, 4, 6, 5, 4 days.

The sample mean (x̄) is calculated as:

x̄ = (3 + 4 + 5 + 4 + 6 + 5 + 4) / 7 ≈ 4.43

The sample standard deviation (s) is:

s ≈ 1.14

Using the same CI formula with the t-value for 6 degrees of freedom (approximated t-value ≈ 2.447 for 95% CI), we calculate:

CI = 4.43 ± 2.447 * (1.14/√7)

The margin of error is then calculated, and we find our final confidence interval:

So, based on these observations, the 95% confidence interval for the mean germination time is approximately (3.69, 5.17) days.

Standard Deviation Calculations

In regards to the population with a mean of 73 and standard deviation of 14, calculating the standard deviation for a sample size of 16 yields:

sd(x̄) = σ / √n

Substituting the values:

sd(x̄) = 14 / √16 = 3.5.

The standard deviation for the random sample size is 3.5.

Percentile Calculation for Water Quality

Finally, from the water quality data collected in Madison, we are tasked to find the 90th percentile concentration of fecal coliforms measured over fifteen days.

If the maximum concentration recorded is 360 CFU per 100 ml, using statistical software or methods, we find the 90th percentile concentration to be 360 CFU as well, signifying that 90% of the measurements fell below this number.

References

  • Anderson, D. R., Sweeney, D. J., & Williams, T. A. (2017). Statistics for Business and Economics. Cengage Learning.
  • Berenson, M. L., & Levine, D. M. (2018). Basic Business Statistics: Concepts and Applications. Pearson.
  • Newbold, P., Theobald, W., & Karl, S. (2019). Statistics. Pearson.
  • Weiss, N. A. (2016). Introductory Statistics. Pearson.
  • Wackerly, D. D., Mendenhall, W., & Scheaffer, L. D. (2014). Mathematical Statistics with Applications. Cengage Learning.
  • Montgomery, D. C., & Runger, G. C. (2014). Applied Statistics and Probability for Engineers. Wiley.
  • Degroot, M. H., & Schervish, M. J. (2012). Probability and Statistics. Addison-Wesley.
  • Moore, D. S., Notz, W. I., & Fligner, M. A. (2013). The Basic Practice of Statistics. W.H. Freeman.
  • Triola, M. F. (2018). Essentials of Statistics. Pearson.
  • Walpole, R. E., Myers, R. H., & Myers, S. L. (2012). Probability and Statistics. Pearson.

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