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What is normal, when it comes to people's body temperatures? A random sample of 130 human body temperatures, provided by Allen Shoemaker in the Journal of Statistical Education, had a mean of \(98.25^{\circ} \mathrm{F}\) and a standard deviation of \(0.73^{\circ} \mathrm{F}\). Does the data indicate that the average body temperature for healthy humans is different from \(98.6^{\circ} \mathrm{F}\), the usual average temperature cited by physicians and others? a. Test using the \(p\) -value approach with \(\alpha=.05\). b. Test using the critical value approach with \(\alpha=.05\). c. Compare the conclusions from parts a and b. Are they the same? d. The 98.6 standard was derived by a German doctor in 1868 , who claimed to have recorded 1 million temperatures in the course of his research. \({ }^{5}\) What conclusions can you draw about his research in light of your conclusions in parts a and b?

Short Answer

Expert verified
Answer: Yes, based on the hypothesis test results, there is a significant difference between the average body temperature of healthy humans and the value of 98.6掳F derived from the 1868 study, as both the p-value approach and the critical value approach led to the rejection of the null hypothesis.

Step by step solution

01

1. State the Null and Alternative Hypotheses

The null hypothesis (H鈧) assumes that there is no difference between the average body temperature and the given value of 98.6掳F. The alternative hypothesis (H鈧) assumes that there is a difference: H鈧: 渭 = 98.6掳F H鈧: 渭 鈮 98.6掳F
02

2. Choose the Significance Level

The significance level, 伪, is already given as 0.05 for both tests (parts a and b).
03

3. Calculate the Test Statistic

We will use a t-test (since the population standard deviation is unknown) for a one-sample hypothesis test: t = (sample mean - hypothesized mean) / (sample standard deviation / 鈭歿sample size}) t = (98.25 - 98.6) / (0.73 / 鈭130) t 鈮 -7.27
04

4a. Calculate the p-value

In part a, we use the p-value approach. Using a two-tailed test with 129 degrees of freedom (since n = 130), we compare the calculated t-value of -7.27 to the t-distribution: p-value = 2 * P(T 鈮 -7.27) = 2 * P(T 鈮 7.27) The p-value is extremely small and can be considered almost negligible.
05

5a. Compare the p-value to the Significance Level

Since the p-value is extremely small and less than the significance level (伪 = 0.05), we reject the null hypothesis (H鈧).
06

4b. Determine the Critical Value

In part b, we use the critical value approach. Again, we use a two-tailed test with 129 degrees of freedom and a 0.05 significance level: critical t-value = 卤 t(伪/2, 129) At 129 DF and 伪 = 0.05, the critical t-value is approximately 卤1.98.
07

5b. Compare the Test Statistic to the Critical Value

Since our calculated test statistic (t 鈮 -7.27) is less than the lower critical value of the two-tailed t-test (-1.98), we reject the null hypothesis (H鈧).
08

6. Compare the Conclusions from Both Approaches

Comparing the conclusions from both approaches, we see that they both lead to the same conclusion: reject the null hypothesis (H鈧). This means that the average body temperature for healthy humans is different from 98.6掳F.
09

7. Discuss the Implications on the Original Research

Our conclusions indicate that the average body temperature for healthy humans is different from the 98.6掳F value derived by the German doctor in 1868. This could mean that his research may have had sampling biases, measurement errors or other factors that led to the determination of 98.6掳F as the standard average body temperature. Additionally, advances in technology have improved the accuracy of temperature measurement since the original research was conducted. Therefore, his claim of 1 million temperatures recorded might not be an accurate representation of the true average body temperature for healthy humans.

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Key Concepts

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

P-Value Approach in Hypothesis Testing
When engaged in hypothesis testing in statistics, the p-value approach is a powerful tool used to decide whether to reject the null hypothesis, which represents the default position or a standard to be tested against.

The p-value represents the probability of observing a test statistic as extreme as, or more extreme than, the actual result, assuming the null hypothesis is true. If this calculated probability is less than the chosen significance level, we conclude that such extreme results are improbable under the null assumption and thus reject the null hypothesis.

In the exercise, the mean body temperature from the sample was compared to the established norm of 98.6掳F. The very low p-value obtained indicated strong evidence against the null hypothesis, resulting in its rejection. Thus, it suggests that the true average body temperature could indeed be different from 98.6掳F.
Critical Value Approach in Hypothesis Testing
An alternative to the p-value approach is the critical value approach. This methodology involves determining a threshold or 'critical value' from the relevant statistical distribution, against which the test statistic will be compared.

In this case, the test-statistic is derived from sample data and is used to assess the plausibility of the null hypothesis. If the test-statistic falls into the critical region determined by the significance level (typically beyond the critical values), the null hypothesis is rejected.

When the one-sample t-test was performed for the body temperature data, the test statistic was much lower than the critical t-value on the negative side. Therefore, by this approach also, the null hypothesis was rejected, aligning the conclusion with that of the p-value approach.
One-Sample T-Test
The one-sample t-test is a statistical method used to compare the mean of a sample to a known value or theoretical expectation鈥攊n this case, the average human body temperature.

The formula for the test statistic in a one-sample t-test is: \[ t = \frac{\text{sample mean} - \text{hypothesized mean}}{\text{sample standard deviation} / \sqrt{\text{sample size}}} \]
For the given exercise, the t-test was used because the population standard deviation was unknown. The large absolute value of the t-statistic obtained indicated substantial deviation from the null hypothesis. This test provides a way to assess if the sample mean significantly differs from the hypothesized mean, which forms the crux of the hypothesis testing exercise.
Significance Level
In hypothesis testing, the significance level, denoted by \( \alpha \), acts as a threshold for making decisions about the null hypothesis. It reflects the risk one is willing to take of incorrectly rejecting a true null hypothesis, a mistake known as Type I error.

The commonly used significance level is 0.05 or 5%, suggesting a 5% risk of concluding a difference when there is none. Both the p-value and critical value approaches in statistics aim to determine whether the evidence is sufficient to reject the null hypothesis at the chosen significance level. Thus, \( \alpha \) represents a balance between sensitivity and specificity in hypothesis tests and is predetermined before the test is conducted to avoid bias.

In the provided example, the significance level set at 0.05 was foundational for determining the rejection of the null hypothesis across both approaches. This level indicates that we demand high evidence before we can reject the null hypothesis and assert that the true average body temperature differs from the traditionally accepted 98.6掳F.

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Most popular questions from this chapter

A random sample of 100 observations from a quantitative population produced a sample mean of 26.8 and a sample standard deviation of \(6.5 .\) Use the \(p\) -value approach to determine whether the population mean is different from \(28 .\) Explain your conclusions.

Refer to Exercise 30 (Section 8.3\()\) and the collection of water samples to estimate the mean acidity (in \(\mathrm{pH}\) ) of rainfalls. Remember that the \(\mathrm{pH}\) for pure rain falling through clean air is approximately 5.7 . The sample of \(n=40\) rainfalls produced \(\mathrm{pH}\) readings with \(\bar{x}=3.7\) and \(s=.5 .\) Do the data provide sufficient evidence to indicate that the mean \(\mathrm{pH}\) for rainfalls is more acidic \(\left(H_{\mathrm{a}}: \mu<5.7 \mathrm{pH}\right)\) than pure rainwater? Test using \(\alpha=.05 .\) Note that this inference is appropriate only for the area in which the rainwater specimens were collected.

In a head-to-head taste test of storebrand foods versus national brands, Consumer Reports found that it was hard to find a taste difference in the two. \(^{16}\) If the national brand is indeed better than the store brand, it should be judged as better more than \(50 \%\) of the time. a. State the null and alternative hypothesis to be tested. Is this a one- or a two-tailed test? b. Suppose that, of the 35 food categories used for the taste test, the national brand was found to be better than the store brand in eight categories. Use this information to test the hypothesis in part a with \(\alpha=.01 .\) What practical conclusions can you draw from the results?

An experiment was planned to compare the mean time (in days) to recover from a common cold for people given a daily dose of 4 milligrams (mg) of vitamin C versus those who were not. Suppose that 35 adults were randomly selected for each treatment category and that the mean recovery times and standard deviations for the two groups were as follows: $$ \begin{array}{lcc} \hline & \begin{array}{l} \text { No Vitamin } \\ \text { Supplement } \end{array} & \begin{array}{c} 4 \mathrm{mg} \\ \text { Vitamin C } \end{array} \\ \hline \text { Sample Size } & 35 & 35 \\ \text { Sample Mean } & 6.9 & 5.8 \\ \text { Sample Standard Deviation } & 2.9 & 1.2 \end{array} $$ a. If you want to show that the use of vitamin \(\mathrm{C}\) reduces the mean time to recover from a common cold, give the null and alternative hypotheses for the test. Is this a one- or a two-tailed test? b. Conduct the statistical test of the null hypothesis in part a and state your conclusion. Test using \(\alpha=.05\)

A new variety of pearl millet is expected to provide an increased yield over the variety presently in use which is about 70 bushels per acre. The new variety of millet produced an average yield of \(\bar{x}=77\) bushels per acre with a standard deviation of \(s=12.6\) bushels based on 40 one-acre yields. Use this information to answer the questions in Exercises \(11-12 .\) Find the value of the test statistic for testing the hypotheses that the new variety will increase yield. Is the value of the test statistic likely, assuming \(H_{0}\) is true?

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