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Total blood volume (in ml) per body weight (in \(\mathrm{kg}\) ) is important in medical research. For healthy adults, the red blood cell volume mean is about \(\mu=28 \mathrm{ml} / \mathrm{kg}\) (Reference: Laboratory and Diagnostic Tests by F. Fischbach). Red blood cell volume that is too low or too high can indicate a medical problem (see reference). Suppose that Roger has had seven blood tests, and the red blood cell volumes were \(\begin{array}{lllllll}32 & 25 & 41 & 35 & 30 & 37 & 29\end{array}\) The sample mean is \(\bar{x} \approx 32.7 \mathrm{ml} / \mathrm{kg} .\) Let \(x\) be a random variable that represents Roger's red blood cell volume. Assume that \(x\) has a normal distribution and \(\sigma=4.75 .\) Do the data indicate that Roger's red blood cell volume is different (either way) from \(\mu=28 \mathrm{ml} / \mathrm{kg}\) ? Use a \(0.01\) level of significance.

Short Answer

Expert verified
Roger's red blood cell volume is significantly different from the mean.

Step by step solution

01

State the Hypotheses

We need to test if Roger's blood cell volume is significantly different from the mean of healthy adults. Therefore, the null hypothesis is \( H_0: \mu = 28 \ \mathrm{ml/kg} \) and the alternative hypothesis is \( H_a: \mu eq 28 \ \mathrm{ml/kg} \).
02

Define the Test Statistic

We will use the Z-test because the population standard deviation \(\sigma\) is known. The test statistic is given by the formula \( Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \), where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.
03

Calculate the Test Statistic

Substitute the given values into the formula: \( \bar{x} = 32.7 \), \( \mu = 28 \), \( \sigma = 4.75 \), and \( n = 7 \). The calculation is as follows:\[Z = \frac{32.7 - 28}{\frac{4.75}{\sqrt{7}}} \approx \frac{4.7}{1.796} \approx 2.62\]
04

Determine the Critical Value

Since we are conducting a two-tailed test with a significance level of \(\alpha = 0.01\), divide the significance level by 2, getting \(0.005\) in each tail. The critical z-value for \(\alpha = 0.005\) in each tail is \(\pm 2.576\).
05

Make a Decision

The calculated test statistic \(Z \approx 2.62\) is greater than the critical value \(2.576\). Therefore, we reject the null hypothesis \(H_0\).
06

State the Conclusion

There is sufficient evidence at the \(0.01\) level of significance to conclude that Roger's red blood cell volume is significantly different from the mean red blood cell volume for healthy adults.

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Key Concepts

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

Z-test
In statistical analysis, a Z-test is an important tool used to determine whether there is a significant difference between a sample mean and a population mean. This is particularly useful when the population standard deviation is known, which in our case of Roger's blood tests, it's given as \(\sigma = 4.75\). The formula for the Z-test statistic is \( Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \), where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, and \( n \) is the sample size.

For Roger's case, the sample mean of his blood cell volumes is approximately 32.7 ml/kg. Using the Z-test helps us decide if this sample mean is significantly different from the population mean of 28 ml/kg for healthy adults. With the Z-test, we calculated the Z-score to be approximately 2.62. This score tells us how many standard deviations Roger's sample mean is above the population mean. When compared with critical values from the Z-distribution, we can make conclusions about the significance of our findings. The test is crucial as it helps identify if the observed difference could be due to random sampling variability or if it's statistically significant.
Normal Distribution
The concept of a normal distribution is fundamental to statistics and is often used in hypothesis testing, like the Z-test applied here. A normal distribution, also known as a Gaussian distribution, is a continuous probability distribution that is symmetrical around its mean, making it look like a bell-shaped curve.

It's important because many real-world variables are distributed normally, which allows statisticians to make inferences about populations. In Roger's blood test analysis, we assume his red blood cell volume follows a normal distribution. So, we can confidently use the Z-test, which relies on the data being normally distributed. The properties of a normal distribution, such as its mean being the center of symmetry and about 68% of the data falling within one standard deviation of the mean, allow us to calculate probabilities and critical values accurately.

Understanding the assumptions of a normal distribution helps clarify why and when specific statistical tests are appropriate. In many testing conditions, particularly with large sample sizes, the Central Limit Theorem supports that sampling distributions of the sample mean tend to be normal, making normal distribution assumptions very practical.
Significance Level
The significance level, denoted by \(\alpha\), is a threshold set by the researcher to decide when to reject the null hypothesis \(H_0\). It essentially sets the probability of making a Type I error, which is rejecting the null hypothesis when it is true. In our analysis of Roger's blood cell volume, we use a significance level of 0.01.

A significance level of 0.01 indicates that there is only a 1% risk of concluding that a difference exists when there is none. This is a stringent criterion, meaning we're demanding strong evidence to reject the null hypothesis. This level is appropriate in medical testing where the cost of an incorrect decision might be high.

During the test, the Z-score calculated was 2.62, which we compared against the critical values aligned with our chosen \(\alpha\) of 0.01. Our critical value was determined to be \(\pm 2.576\) for a two-tailed test setup, and since our test statistic exceeded these critical values, we rejected the null hypothesis. Choosing a significance level is a crucial part of hypothesis testing as it reflects the researcher's tolerance for error and impacts the reliability of the test's conclusions.

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Most popular questions from this chapter

USA Today reported that the state with the longest mean life span is Hawaii, where the population mean life span is 77 years. A random sample of 20 obituary notices in the Honolulu Advertizer gave the following information about life span (in years) of Honolulu residents: \(\begin{array}{llllllllll}72 & 68 & 81 & 93 & 56 & 19 & 78 & 94 & 83 & 84 \\\ 77 & 69 & 85 & 97 & 75 & 71 & 86 & 47 & 66 & 27\end{array}\) i. Use a calculator with mean and standard deviation keys to verify that \(\bar{x}=71.4\) years and \(s=20.65\) years. ii. Assuming that life span in Honolulu is approximately normally distributed, does this information indicate that the population mean life span for Honolulu residents is less than 77 years? Use a \(5 \%\) level of significance.

This problem is based on information regarding productivity in leading Silicon Valley companies (see reference in Problem 25). In large corporations, an "intimidator" is an employee who tries to stop communication, sometimes sabotages others, and, above all, likes to listen to him- or herself talk. Let \(x_{1}\) be a random variable representing productive hours per week lost by peer employees of an intimidator. \(\begin{array}{llllllll}x_{1}: & 8 & 3 & 6 & 2 & 2 & 5 & 2\end{array}\) A "stressor" is an employee with a hot temper that leads to unproductive tantrums in corporate society. Let \(x_{2}\) be a random variable representing productive hours per week lost by peer employees of a stressor. \(\begin{array}{lllllllll}x_{2}: & 3 & 3 & 10 & 7 & 6 & 2 & 5 & 8\end{array}\) i. Use a calculator with mean and standard deviation keys to verify that \(\bar{x}_{1}=4.00, s_{1}=2.38, \bar{x}_{2}=5.5\), and \(s_{2}=2.78 .\) ii. Assuming the variables \(x_{1}\) and \(x_{2}\) are independent, do the data indicate that the population mean time lost due to stressors is greater than the population mean time lost due to intimidators? Use a \(5 \%\) level of significance. (Assume the population distributions of time lost due to intimidators and time lost due to stressors are each mound-shaped and symmetric.)

A random sample of 25 values is drawn from a mound-shaped and symmetric distribution. The sample mean is 10 and the sample standard deviation is \(2 .\) Use a level of significance of \(0.05\) to conduct a two-tailed test of the claim that the population mean is \(9.5\). (a) Cbeck Requirements Is it appropriate to use a Student's \(t\) distribution? Explain. How many degrees of freedom do we use? (b) What are the hypotheses? (c) Compute the sample test statistic \(t .\) (d) Estimate the \(P\) -value for the test. (e) Do we reject or fail to reject \(H_{0} ?\) (f) Interpret the results.

Let \(x\) be a random variable that represents red blood cell (RBC) count in millions of cells per cubic millimeter of whole blood. Then \(x\) has a distribution that is approximately normal. For the population of healthy female adults, the mean of the \(x\) distribution is about \(4.8\) (based on information from Diagnostic Tests with Nursing Implications, Springhouse Corporation). Suppose that a female patient has taken six laboratory blood tests over the past several months and that the \(\mathrm{RBC}\) count data sent to the patient's doctor are \(\begin{array}{llllll}4.9 & 4.2 & 4.5 & 4.1 & 4.4 & 4.3\end{array}\) i. Use a calculator with sample mean and sample standard deviation keys to verify that \(\bar{x}=4.40\) and \(s \approx 0.28\). ii. Do the given data indicate that the population mean \(\mathrm{RBC}\) count for this patient is lower than \(4.8\) ? Use \(\alpha=0.05\).

Is fishing better from a boat or from the shore? Pyramid Lake is located on the Paiute Indian Reservation in Nevada. Presidents, movie stars, and people who just want to catch fish go to Pyramid Lake for really large cutthroat trout. Let row \(B\) represent hours per fish caught fishing from the shore, and let row \(A\) represent hours per fish caught using a boat. The following data are paired by month from October through April (Source: Pyramid Lake Fisheries, Paiute Reservation, Nevada). $$ \begin{array}{l|ccccccc} \hline & \text { Oct. } & \text { Nov. } & \text { Dec. } & \text { Jan. } & \text { Feb. } & \text { March } & \text { April } \\ \hline \text { B: Shore } & 1.6 & 1.8 & 2.0 & 3.2 & 3.9 & 3.6 & 3.3 \\ \hline \text { A: Boat } & 1.5 & 1.4 & 1.6 & 2.2 & 3.3 & 3.0 & 3.8 \\ \hline \end{array} $$ Use a \(1 \%\) level of significance to test if there is a difference in the population mean hours per fish caught using a boat compared with fishing from the shore.

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