We Want To Compare The Mean Hospital Stay In Pennsylvania

We Want To Compare The Mean Hospital Stay In A Pennsylvania Hospital A

We want to compare the mean hospital stay in a Pennsylvania hospital and the mean hospital stay in a typical U.S. hospital. It is unknown whether the stay in a Pennsylvania hospital is longer or shorter than a stay in a typical U.S. hospital. Assume the duration of a hospital stay in the U.S. is normally distributed and the mean is 11.5 days. So, is the mean hospital stay in Pennsylvania different from the mean hospital stay in a typical U.S. hospital? a) Let's start by stating the null and alternative hypotheses. Also, what is your test statistic and why? b) Continuing with the hospital data, what were your test statistic results?

Paper For Above instruction

Introduction

The comparison of hospital stay durations across different regions provides valuable insights into healthcare efficiency, patient recovery rates, and resource utilization. Specifically, understanding whether the mean hospital stay in a Pennsylvania hospital deviates significantly from the national average can inform healthcare policies and hospital management practices. This analysis employs inferential statistical methods to determine if there is a statistically significant difference between these two means, under the assumption that the U.S. hospital stay duration is normally distributed with a known mean of 11.5 days.

Formulating Hypotheses

The first step in the hypothesis testing process is to establish the statistical hypotheses. The null hypothesis (H0) assumes that there is no difference between the mean hospital stay in the Pennsylvania hospital and the national average. This can be mathematically expressed as H0: μ = 11.5 days. Conversely, the alternative hypothesis (H1) posits that the mean hospital stay in the Pennsylvania hospital differs from the national average, expressed as H1: μ ≠ 11.5 days. These hypotheses set the stage for a two-tailed hypothesis test, as the question seeks to identify any difference, regardless of direction.

Selection of Test Statistic

The choice of test statistic depends on the nature of the data and the known population parameters. Here, the population mean for U.S. hospitals is known as 11.5 days, and the data from the Pennsylvania hospital is assumed to be approximately normally distributed. If the standard deviation (σ) of hospital stays in Pennsylvania is known or the sample size is large enough (generally n > 30), a Z-test is appropriate. Otherwise, with an unknown standard deviation and a smaller sample, a t-test would be preferred.

Given that the problem statement suggests assuming normality for the U.S. distribution and does not specify the population standard deviation, we generally proceed with a t-test if the sample standard deviation (s) is used. The test statistic is calculated as:

t = (x̄ - μ₀) / (s / √n)

where x̄ is the sample mean from the Pennsylvania hospital, μ₀ is the hypothesized mean (11.5 days), s is the sample standard deviation, and n is the sample size.

This test statistic measures how many standard errors the sample mean is away from the hypothesized population mean, providing the basis for determining statistical significance.

Results of the Test Statistic

Suppose the hospital data collection yielded a sample mean (x̄) of 12 days with a sample standard deviation (s) of 3 days, based on a sample size of 40 patients (n=40). The calculations would proceed as follows:

t = (12 - 11.5) / (3 / √40) = 0.5 / (3 / 6.3246) ≈ 0.5 / 0.4743 ≈ 1.055

Referring to the t-distribution table with 39 degrees of freedom (n-1), a t-value of approximately 1.055 corresponds to a p-value greater than 0.2 in a two-tailed test. Since this p-value exceeds common significance levels (0.05 or 0.01), we fail to reject the null hypothesis.

In conclusion, based on the sample data, there is insufficient evidence to suggest that the mean hospital stay in the Pennsylvania hospital differs significantly from the national average of 11.5 days.

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