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Human oral normal body temperature is believed to be \(98.6^{\circ} \mathrm{F},\) but there is evidence that it actually should be \(98.2^{\circ} \mathrm{F}\) [Mackowiak, Wasserman, Steven and Levine, JAMA (1992, Vol. \(268(12),\) pp. \(1578-1580)] .\) From a sample of 52 healthy adults, the mean oral temperature was 98.285 with a standard deviation of 0.625 degrees. (a) What are the null and alternative hypotheses? (b) Test the null hypothesis at \(\alpha=0.05\). (c) How does a \(95 \%\) confidence interval answer the same question?

Short Answer

Expert verified
Reject the null hypothesis; the mean temperature is not 98.6°F. The 95% confidence interval (98.112, 98.458) doesn't include 98.6°F, supporting this conclusion.

Step by step solution

01

Define the Hypotheses

To perform a hypothesis test, we first need to define the null and alternative hypotheses. - The **null hypothesis** (\(H_0\)) states that the mean oral body temperature is still 98.6°F. Thus, \(H_0: \mu = 98.6\). - The **alternative hypothesis** (\(H_a\)) suggests that the mean is not 98.6°F. Thus, \(H_a: \mu eq 98.6\). This is a two-tailed test because we are interested in any deviation from 98.6°F.
02

Find the Test Statistic

We use the sample mean (\(\bar{x} = 98.285\)), the population mean under the null hypothesis (\(\mu = 98.6\)), the sample standard deviation (\(s = 0.625\)), and the sample size (\(n = 52\)) to calculate the test statistic. The formula for the t-statistic is: \[t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \approx \frac{98.285 - 98.6}{0.625 / \sqrt{52}} \]Calculating each part gives: the denominator as approximately 0.0867, leading to a t-value of approximately -3.629.
03

Determine the Critical Value and Decision

With \(\alpha = 0.05\), a two-tailed test, and degrees of freedom \(n-1=51\), we look up the critical t-value from the t-distribution table. The critical t-value is approximately \(\pm2.008\). Our computed test statistic \(t \approx -3.629\) is less than \(-2.008\). Since the test statistic falls in the rejection region, we reject the null hypothesis.
04

Interpret the 95% Confidence Interval

To construct a 95% confidence interval, use the formula:\[\bar{x} \pm t_{\alpha/2} \left(\frac{s}{\sqrt{n}}\right)\]Substitute the values: \[98.285 \pm 2.008 \left(\frac{0.625}{\sqrt{52}}\right)\]The confidence interval for the mean temperature becomes approximately (98.112, 98.458). Since 98.6°F is not within this interval, it supports rejecting the null hypothesis.

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Key Concepts

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

Confidence Interval
A confidence interval is a range of values used to estimate a population parameter. In the context of hypothesis testing, it provides an estimated range where we expect the true population mean to fall. The calculation of a confidence interval involves the sample mean, the standard deviation, and the critical t-value related to the selected confidence level.
For instance, in our example, a 95% confidence interval was computed using the following formula: \( \bar{x} \pm t_{\alpha/2} \left(\frac{s}{\sqrt{n}}\right) \).
This includes:
  • \( \bar{x} = 98.285 \), the sample mean temperature.
  • \( s = 0.625 \), the standard deviation of the sample.
  • \( n = 52 \), the sample size.
  • \( t_{\alpha/2} = 2.008 \), the critical value from the t-distribution table at \( \alpha = 0.05 \).
This process resulted in a confidence interval of approximately (98.112, 98.458). Since the supposedly true mean of 98.6°F is outside this range, it provides evidence against the null hypothesis, suggesting that 98.6°F is not a reasonable estimate of the population mean.
t-statistic
The t-statistic is a ratio that comes into play when you want to decide whether to reject a null hypothesis in hypothesis testing. It compares the observed sample mean to the population mean under the null hypothesis, scaling by the variability observed in the sample and the size of the sample.
The t-statistic formula is: \[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \]where:
  • \( \bar{x} \) is the sample mean
  • \( \mu \) is the population mean under the null hypothesis
  • \( s \) is the standard deviation of the sample
  • \( n \) is the sample size
In our example, substituting the values gives us a t-statistic of approximately \(-3.629\). This t-value is then compared to a critical value from the t-distribution with \( n-1 \) degrees of freedom.
The t-statistic tells us how far our sample mean is from the hypothesized population mean, in units of standard error. A high absolute value of the t-statistic indicates a large deviation from the null hypothesis.
Null and Alternative Hypotheses
In hypothesis testing, we work with two main hypotheses: the null hypothesis and the alternative hypothesis. The null hypothesis (\( H_0 \)) represents the default or original assumption, which, in many cases, claims that there is no effect or difference.
In this exercise, the null hypothesis is that the mean oral body temperature is 98.6°F, or \( H_0: \mu = 98.6 \).
The alternative hypothesis (\( H_a \)) reflects the researcher's question or suspicion that contradicts the null hypothesis. It posits that there is an effect or a difference.
For this example, the alternative hypothesis is that the mean is not 98.6°F, written as \( H_a: \mu eq 98.6 \).
These hypotheses help guide the entire statistical test, determining whether the sample data provides sufficient evidence to reject the null hypothesis.
Two-Tailed Test
A two-tailed test is a type of hypothesis test where we examine the possibility of the sample mean being either significantly higher or lower than the hypothesized population mean.
This is in contrast to a one-tailed test where we test the direction of the deviation.
In a two-tailed test, the null hypothesis is tested against an alternative that allows for deviations on both sides.
For the task at hand, the hypotheses \( H_0: \mu = 98.6 \) and \( H_a: \mu eq 98.6 \) mean that we are interested in whether the actual mean temperature significantly deviates from 98.6°F, be it higher or lower.
  • Critical regions are located on both extremes of the distribution.
  • This approach increases the range of conditions under which the null hypothesis can be rejected.
In this way, a two-tailed test provides a more comprehensive assessment for scenarios where deviations may occur in either direction.

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Most popular questions from this chapter

Output from a software package follows: Test of \(m u=99\) vs \(>99\) The assumed standard deviation \(=2.5\) $$\begin{array}{ccccccc}\text { Variable } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } & \mathrm{Z} & \mathrm{P} \\\x & 12 & 100.039 & 2.365 & ? & 1.44 & 0.075 \\\\\hline\end{array}$$ (a) Fill in the missing items. What conclusions would you draw? (b) Is this a one-sided or a two-sided test? (c) If the hypothesis had been \(H_{0}: \mu=98\) versus \(H_{0}: \mu>98\), would you reject the null hypothesis at the 0.05 level of significance? Can you answer this without referring to the normal table? (d) Use the normal table and the preceding data to construct a \(95 \%\) lower bound on the mean. (e) What would the \(P\) -value be if the alternative hypothesis is \(H_{1}: \mu \neq 99 ?\)

For the hypothesis test \(H_{0}: \mu=7\) against \(H_{1}: \mu \neq 7\) with variance unknown and \(n=20\), approximate the \(P\) -value for each of the following test statistics. (a) \(t_{0}=2.05\) (b) \(t_{0}=-1.84\) (c) \(t_{0}=0.4\)

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