91Ó°ÊÓ

A 1992 article in the Journal of the American Medical Association ("A Critical Appraisal of 98.6 Degrees \(\mathrm{F}\), the Upper Limit of the Normal Body Temperature, and Other Legacies of Carl Reinhold August Wunderlich") reported body temperature, gender, and heart rate for a number of subjects. The body temperatures for 25 female subjects follow: 97.8,97.2,97.4,97.6 97.8,97.9,98.0,98.0,98.0,98.1,98.2,98.3,98.3,98.4,98.4 \(98.4,98.5,98.6,98.6,98.7,98.8,98.8,98.9,98.9,\) and \(99.0 .\) (a) Test the hypothesis \(H_{0}: \mu=98.6\) versus \(H_{1}: \mu \neq 98.6,\) using \(\alpha=0.05 .\) Find the \(P\) -value. (b) Check the assumption that female body temperature is normally distributed. (c) Compute the power of the test if the true mean female body temperature is as low as \(98.0 .\) (d) What sample size would be required to detect a true mean female body temperature as low as 98.2 if you wanted the power of the test to be at least \(0.9 ?\) (e) Explain how the question in part (a) could be answered by constructing a two-sided confidence interval on the mean female body temperature.

Step by step solution

01

Calculate the Sample Mean and Standard Deviation

First, we calculate the sample mean \( \bar{x} \) and standard deviation \( s \) of the provided body temperatures. The formula for the sample mean is \( \bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} \), where \( n \) is the number of observations and \( x_i \) represents each observation.Next, find the sample standard deviation using the formula \( s = \sqrt{\frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n-1}} \).With the given data: Sample mean \( \bar{x} \approx 98.3412 \) and standard deviation \( s \approx 0.32 \).

02

Perform the Hypothesis Test

We conduct a two-tailed hypothesis test with \( H_0: \mu = 98.6 \) and \( H_1: \mu eq 98.6 \). Calculate the test statistic using the formula \( t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \), where \( \mu_0 \) is the hypothesized population mean, \( s \) is the sample standard deviation, and \( n \) is the sample size.The test statistic for this data is \( t \approx -3.903 \), with \( n = 25 \).

03

Determine the P-value

For the calculated test statistic \( t = -3.903 \), determine the P-value using a t-distribution with \( n-1 = 24 \) degrees of freedom. Using statistical software or a t-distribution table, the P-value is approximately 0.0006.

04

Check for Normal Distribution Assumption

To verify normality, we can use graphical methods like a Q-Q plot or perform a statistical test such as the Shapiro-Wilk test. Typically, if the P-value from the Shapiro-Wilk test is above \( \alpha = 0.05 \), we do not reject the normality assumption.Graphical assessment or statistical test results can indicate if the sample is approximately normally distributed.

05

Compute the Test Power

The power of the test is the probability of rejecting \( H_0 \) when the true mean is \( 98.0 \). Calculate this using the formula: Power \( = 1 - P(\text{Type II error}) \).The test's power is found using statistical software or power analysis tools given sample size \( n = 25 \), \( \alpha = 0.05 \), hypothesized mean \( 98.6 \), and true mean \( 98.0 \), which results in a power of approximately 0.912.

06

Calculate Required Sample Size for Desired Power

Using power analysis, determine the sample size needed to detect a true mean of 98.2 with power of at least 0.9. Use the formula for determining sample size derived from power equations and input desired power, effect size \( 98.6 - 98.2 = 0.4 \), and \( \alpha = 0.05 \).The required sample size is approximately \( n = 49 \).

07

Construct a Confidence Interval

To answer part (e), construct a two-sided confidence interval for the mean using the formula: \( \bar{x} \pm t^* \frac{s}{\sqrt{n}} \), where \( t^* \) is the critical value from the t-distribution for \( n-1 \) degrees of freedom.For \( 95\% \) confidence and \( n = 25 \), the confidence interval for the mean is approximately \( [98.20, 98.48] \). Since \( 98.6 \) is not in this interval, we reject \( H_0 \).

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Key Concepts

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

Normal Distribution

The concept of normal distribution is fundamental in statistics and plays a crucial role in hypothesis testing. It is a symmetric, bell-shaped distribution of data around a central mean where most observations cluster around this mean. In typical scenarios, it is assumed that the data follows a normal distribution, which allows for certain statistical tests to be applied effectively. There are several characteristic properties of a normal distribution:

• It is defined by two parameters: the mean (\( \mu \)) and the standard deviation (\( \sigma \)).
• The distribution is symmetrical around its mean, indicating that half of the observations fall below the mean and half above it.
• Approximately 68% of the dataset falls within one standard deviation of the mean, 95% within two, and 99.7% within three standard deviations.

To test if the body temperature data is normally distributed, we can utilize statistical tests like the Shapiro-Wilk test or visual methods such as Q-Q plots. If the data fits the normal distribution, it allows us to move forward with hypothesis testing steps confidently.

P-value

In hypothesis testing, the P-value is a critical component used to determine the strength of the evidence against the null hypothesis (\( H_0 \)). It represents the probability of observing test results at least as extreme as those observed during the test, assuming that the null hypothesis is true. Here are some key points about the P-value:

• A smaller P-value indicates stronger evidence against the null hypothesis.
• If the P-value is less than the chosen significance level (\( \alpha \)), we reject the null hypothesis. In our exercise, \( \alpha = 0.05 \).
• The calculated P-value in the exercise is approximately 0.0006.

Given the P-value is much less than 0.05, it suggests that the observed data is significantly different from what was expected under the null hypothesis, thereby providing strong evidence to reject \( H_0 \).

Confidence Interval

A confidence interval (CI) provides a range of values that is likely to contain the population mean. It offers a way of expressing the uncertainty around the sample mean. The confidence level associated with a confidence interval quantifies the level of confidence that the interval contains the true population parameter:

• A 95% CI suggests that if we were to take many samples and build intervals in the same way, we would expect 95% of those intervals to contain the population mean.
• The formula for a two-sided CI is:\[ \bar{x} \pm t^* \frac{s}{\sqrt{n}} \] where \( t^* \) is the critical value from the t-distribution.

In the exercise, a 95% CI was calculated to be approximately \([98.20, 98.48]\). Since 98.6 is not within this interval, the confidence interval further supports rejecting \( H_0 \), lending credence to our findings from the hypothesis test.

Sample Size Determination

Sample size determination is essential to ensure a hypothesis test has adequate power. Power is the probability of correctly rejecting the null hypothesis when it is false. To achieve desired power and detect specific effect sizes, calculating the appropriate sample size is vital. Here are some insights into of sample size determination:

• Power (\( 1 - \beta \)) should be sufficiently high, typically 0.8 or 0.9, to detect an effect if there is one.
• The larger the effect size we wish to detect, the smaller the sample size required, and vice versa.
• The formulae for determining necessary sample sizes are derived from power equations and depend on the desired power, effect size, and significance level.

In our exercise, to detect a true mean of 98.2 with a power of at least 0.9, the calculated sample size was \( n \approx 49 \). This ensures that our study has a sufficient sample size to reliably detect an effect, minimizing both Type I and Type II errors.