/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 The postanesthesia care area (re... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 38

The postanesthesia care area (recovery room) at St. Luke's Hospital in Maumee, Ohio, was recently enlarged The hope was that with the enlargement the mean number of patients per day would be more than \(25 .\) A random sample of 15 days revealed the following numbers of patients $$\begin{array}{|lllllllllllllll|}\hline 25 & 27 & 25 & 26 & 25 & 28 & 28 & 27 & 24 & 26 & 25 & 29 & 25 & 27 & 24 \\\\\hline\end{array}$$ At the .01 significance level, can we conclude that the mean number of patients per day is more than \(25 ?\) Estimate the \(p\) value and interpret it.

Short Answer

Expert verified
Reject the null hypothesis; the p-value indicates the mean is significantly greater than 25.

Step by step solution

01

State the Hypotheses

To determine if the mean number of patients per day is more than 25, we need to set up the null and alternative hypotheses. The null hypothesis (H_0) is that the mean number of patients per day is 25: \[H_0: \mu = 25\]The alternative hypothesis (H_a) is that the mean number of patients per day is greater than 25: \[H_a: \mu > 25\]
02

Calculate the Sample Mean and Standard Deviation

Calculate the sample mean and standard deviation from the given data.Sample data: 25, 27, 25, 26, 25, 28, 28, 27, 24, 26, 25, 29, 25, 27, 24Sample mean (\bar{x}) is given by:\[\bar{x} = \frac{25 + 27 + 25 + 26 + 25 + 28 + 28 + 27 + 24 + 26 + 25 + 29 + 25 + 27 + 24}{15} = 26\]To compute the sample standard deviation (s):\[s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}\]
03

Calculate the Test Statistic

We use the t-test for the mean since the sample size is small (n=15). The test statistic (T) is given by:\[T = \frac{\bar{x} - \mu}{s/\sqrt{n}}\]Where \(\bar{x} = 26\), \(\mu = 25\), \(n = 15\), and you have calculated \(s\) from the previous step.
04

Determine the Critical Value and Compare

The critical value for a one-tailed test at the 0.01 significance level with 14 degrees of freedom (n-1) can be found using a t-table. Locate the critical value \(t_{0.01, 14}\).If \(|T|\) from the previous step is greater than \(t_{0.01, 14}\), reject the null hypothesis.
05

Estimate the p-value and Interpret

Using the t-distribution calculator or table, determine the p-value associated with the calculated test statistic. Interpretation: - If the p-value is less than 0.01, reject the null hypothesis and conclude the mean number of patients per day is significantly greater than 25. - If the p-value is greater than 0.01, fail to reject the null hypothesis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

t-test
When you have a small sample size and want to test if your sample mean significantly differs from a known population mean, the t-test comes into play. In our exercise, we performed a one-sample t-test to see if the mean number of patients was greater than 25.
  • The one-sample t-test is ideal when the sample size is small (usually below 30).
  • It compares the sample mean to a theoretical population mean.
Using the formula for the t-test:\[T = \frac{\bar{x} - \mu}{s/\sqrt{n}}\]where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size, we computed the t-statistic. The t-statistic informs us how far and in what direction our sample mean deviates from the null hypothesis mean under the standard error measure.
p-value
The p-value provides the probability of obtaining test results at least as extreme as those observed, under the assumption that the null hypothesis is correct. In other words, it helps measure the evidence against the null hypothesis.
  • A smaller p-value indicates stronger evidence against the null hypothesis.
  • It quantifies the risk of making a wrong decision (Type I error) when rejecting the null hypothesis.
In our case, we estimate the p-value using the calculated t-statistic. If this value is less than the predetermined significance level, we can reject the null hypothesis. The idea is straightforward: any p-value smaller than your chosen significance threshold implies that the observed data is sufficiently unexpected under the null hypothesis.
sample mean
The sample mean is a key statistic used in hypothesis testing as it estimates the average value of a dataset. In our exercise involving St. Luke's Hospital, the sample mean represents the average number of patients per day taken from the sample of 15 days.
  • To calculate the sample mean, sum up all the sample values and divide by the number of samples.
  • It serves as an estimator for the population mean.
In the formula:\[\bar{x} = \frac{\sum x_i}{n}\]where \(x_i\) represents each value in the dataset and \(n\) is the sample size, 26 was our calculated sample mean. This value was then used in the t-test to determine if it significantly differed from the hypothesized population mean of 25 patients.
significance level
The significance level, denoted as \(\alpha\), is a threshold set by the researcher that defines how extreme observed data must be for us to reject the null hypothesis.
  • A common significance level is 0.05, but in our example, we use 0.01, indicating a more stringent criterion for significance.
  • Essentially, it reflects how willing we are to risk making a Type I error鈥攔ejecting a true null hypothesis.
Using a 0.01 significance level means we're asking for stronger evidence before concluding that the mean number of patients is greater than 25. If our p-value is less than 0.01, it suggests that the result is statistically significant; we can then confidently reject the null hypothesis and assert that the enlargement had the desired effect in increasing patient numbers.

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Most popular questions from this chapter

A spark plug manufacturer claimed that its plugs have a mean life in excess of 22,100 miles. Assume the life of the spark plugs follows the normal distribution. A fleet owner purchased a large number of sets. A sample of 18 sets revealed that the mean life was 23,400 miles and the standard deviation was 1,500 miles. Is there enough evidence to substantiate the manufacturer's claim at the .05 significance level?

Research at the University of Toledo indicates that 50 percent of students change their major area of study after their first year in a program. A random sample of 100 students in the College of Business revealed that 48 had changed their major area of study after their first year of the program. Has there been a significant decrease in the proportion of students who change their major after the first year in this program? Test at the .05 level of significance.

According to the local union president, the mean income of plumbers in the Salt Lake City area follows the normal probability distribution with a mean of \(\$ 45,000\) and a standard deviation of \(\$ 3,000 .\) A recent investigative reporter for KYAK TV found, for a sample of 120 plumbers, the mean income was \(\$ 45,500\). At the .10 significance level, is it reasonable to conclude that the mean income is not equal to \(\$ 45,000 ?\) Determine the \(p\) value.

(a) Is this a one- or two-tailed test? (b) What is the decision rule? (c) What is the value of the test statistic? (d) What is your decision regarding \(H_{0} ?\) (e) What is the \(p\) -value? Interpret it. A sample of 64 observations is selected from a normal population. The sample mean is \(215,\) and the population standard deviation is \(15 .\) Conduct the following test of hypothesis using the .03 significance level.$$\begin{array}{l}H_{0} ; \mu \geq 220 \\\H_{1}: \mu<220\end{array}$$

(a) Is this a one- or two-tailed test? (b) What is the decision rule? (c) What is the value of the test statistic? (d) What is your decision regarding \(H_{0} ?\) (e) What is the \(p\) -value? Interpret it. A sample of 36 observations is selected from a normal population. The sample mean is \(21,\) and the population standard deviation is \(5 .\) Conduct the following test of hypothesis using the .05 significance level. $$\begin{array}{l}H_{0}: \mu \leq 20 \\\H_{1}: \mu>20\end{array}$$
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