Help Make A Decision About Expansion Plans

To Help Make A Decision About Expansion Plans The

To Help Make A Decision About Expansion Plans The

To help make a decision about expansion plans, the director of a music company needs to know how many compact discs teenagers buy annually. Accordingly, she commissions you, a recent employee, to conduct a survey. Suppose that you randomly selected 250 teenagers and asked each to report the number of CDs purchased in the previous 12 months. You found that the sample mean is 4.26 CDs and the sample standard deviation is 3 CDs.

Construct a 95% confidence interval around the sample mean. Be sure to show whether you use the t- or z- value, and what value you use. Interpret the confidence interval in a single sentence. In another sentence, state why you used either the “t” or the “z”.

Paper For Above instruction

The objective of this analysis is to estimate the average number of compact discs purchased annually by teenagers, based on a sample of 250 teenagers, with the aim of informing the company's expansion strategy. Using this sample data, we will construct a 95% confidence interval around the sample mean to provide a range within which the true population mean likely lies, with a specified level of confidence.

Given the sample size of 250, which is quite large, and the standard deviation of 3 CDs, the appropriate statistical approach involves calculating a confidence interval using the z-distribution. The Central Limit Theorem supports this choice because with large samples, the sampling distribution of the mean approaches normality, even if the underlying distribution is not normal.

First, we identify the relevant parameters: the sample mean (x̄) is 4.26 CDs, and the sample standard deviation (s) is 3 CDs. The sample size (n) is 250. The standard error of the mean (SE) is computed as:

SE = s / √n = 3 / √250 ≈ 3 / 15.811 ≈ 0.1897

Next, for a 95% confidence level, the critical value from the standard normal distribution (z-value) is approximately 1.96. The confidence interval (CI) is calculated as:

CI = x̄ ± z SE = 4.26 ± 1.96 0.1897 ≈ 4.26 ± 0.371

Therefore, the 95% confidence interval is approximately (3.89, 4.63).

This interval suggests that we can be 95% confident that the true average number of CDs purchased annually by teenagers lies between approximately 3.89 and 4.63 CDs.

The reason for using the z-distribution instead of the t-distribution is that, with a large sample size of 250, the sampling distribution of the mean can be approximated by a normal distribution according to the Central Limit Theorem. The z-distribution provides a valid approximation in this context because the sample size is sufficiently large, and the population standard deviation is unknown but estimated by the sample standard deviation, which is acceptable here due to the large sample size.

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