We Test If The Mean Rent Of Downtown Pittsburgh One Bedroom

Question

We Test If The Mean Rent Of All Downtown Pittsburgh One Bedroom Apartm We test if the mean rent of all downtown Pittsburgh one-bedroom apartments is different from the mean rent of all non-downtown one-bedroom apartments by examining a random sample of downtown one-bedroom apartments and a random sample of non-downtown one-bedroom apartments. A) One-way Analysis of Variance (ANOVA) B) Paired t-test with a two-sided alternative C) Regression D) Two sample proportion with a two-sided alternative E) T-test about a mean with a two-sided alternative F) Chi-square test of independence G) Two sample t-test with a two-sided alternative 1. To help make a decision about expansion plans, the director of a music company needs to know how many compact disc 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”.

Dr. Jack HW Helper · Accepted Answer

To address the query regarding the average number of compact discs (CDs) purchased annually by teenagers, a statistical analysis was conducted based on a random sample of 250 teenagers. The sample yielded a mean of 4.26 CDs and a standard deviation of 3 CDs. This analysis involves constructing a 95% confidence interval (CI) to estimate the true population mean, leveraging the statistical properties associated with large sample sizes. Constructing the Confidence Interval Given the sample size (n = 250) is large, we can confidently apply the z-approach for constructing the confidence interval. The general formula for a confidence interval for the population mean when the population standard deviation is unknown but the sample size is large involves the z-distribution, which approximates the normal distribution. The formula for the 95% confidence interval is: CI = x̄ ± z* (s / √n) Where: x̄ = 4.26 (sample mean) s = 3 (sample standard deviation) n = 250 (sample size) z* = 1.96 (z-value corresponding to a 95% confidence level) Calculations Calculating the standard error: SE = s / √n = 3 / √250 ≈ 3 / 15.81 ≈ 0.19 Therefore, the confidence interval becomes: CI = 4.26 ± 1.96 × 0.19 ≈ 4.26 ± 0.37 This yields: Lower limit ≈ 3.89 Upper limit ≈ 4.63 Interpretation of the Confidence Interval We are 95% confident that the true average number of CDs purchased annually by teenagers falls between approximately 3.89 and 4.63 CDs. Rationale for Using the z-Value The use of the z-value is justified in this context because the sample size is sufficiently large (n = 250), which satisfies the conditions for applying the Central Limit Theorem. This allows the approximation of the sampling distribution of the mean to a normal distribution, even if the population distribution is not perfectly normal. Furthermore, standard practice in large-sample scenarios involves using the z-distribution for constructing confidence intervals when the population variance is unknown but estimated from the sa

We Test If The Mean Rent Of Downtown Pittsburgh One Bedroom

We Test If The Mean Rent Of All Downtown Pittsburgh One Bedroom Apartm

Paper For Above instruction

Constructing the Confidence Interval

Calculations

Interpretation of the Confidence Interval

Rationale for Using the z-Value

Conclusion

References