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Output from a software package is given below: One-Sample Z: Test of \(m u=14.5 \mathrm{vs}>14.5\) The assumed standard deviation \(=1.1\) $$ \begin{array}{lccccccc} \text { Variable } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } & \mathrm{Z} & \mathrm{P} \\ \mathrm{x} & 16 & 15.016 & 1.015 & ? & ? & ? \end{array} $$ (a) Fill in the missing items. What conclusions would you draw? (b) Is this a one-sided or a two-sided test? (c) Use the normal table and the above data to construct a \(95 \%\) lower bound on the mean. (d) What would the \(P\) -value be if the alternative hypothesis is \(H_{1}: \mu \neq 14.5 ?\)

Short Answer

Expert verified
(a) Reject null hypothesis; mean > 14.5. P = 0.0304. (b) One-sided test. (c) Lower bound: 14.5636. (d) Two-sided P-value = 0.0608.

Step by step solution

01

Calculate Standard Error of Mean (SE Mean)

To find the standard error of the mean, use the formula: \( SE = \frac{\sigma}{\sqrt{n}} \). Here, \( \sigma = 1.1 \) (standard deviation) and \( n = 16 \) (sample size). Therefore, \( SE = \frac{1.1}{\sqrt{16}} = \frac{1.1}{4} = 0.275 \).
02

Calculate Z value

To calculate the Z value for the given test, use the formula: \( Z = \frac{\bar{x} - \mu}{SE} \). Here, \( \bar{x} = 15.016 \), \( \mu = 14.5 \), and \( SE = 0.275 \). Thus, \[ Z = \frac{15.016 - 14.5}{0.275} \approx \frac{0.516}{0.275} \approx 1.876 \].
03

Determine P-value

Using the Z value of approximately 1.876, refer to a Z-table or standard normal distribution table to find the corresponding P-value for a one-tailed test. For \( Z = 1.876 \), the P-value is approximately 0.0304.
04

Conclusion for Part (a)

Since the P-value (0.0304) is less than the common significance level of 0.05, reject the null hypothesis. There is sufficient evidence to support that the mean is greater than 14.5.
05

Determine Test Type for Part (b)

The test compares \( \mu = 14.5 \) vs. \( \mu > 14.5 \), which indicates it is a one-sided test.
06

Construct 95% Lower Bound on the Mean for Part (c)

To construct a 95% lower bound, use \( \bar{x} - Z_{\alpha} \times SE \) where \( Z_{\alpha} \) for 95% one-tailed is approximately 1.645. So, \( 15.016 - 1.645 \times 0.275 = 15.016 - 0.452375 \approx 14.5636 \). The lower bound is approximately 14.5636.
07

Calculate P-value for Two-Sided Test for Part (d)

For a two-sided test with the same Z value (1.876), multiply the previously found one-tailed P-value by 2. Thus, \( P = 2 \times 0.0304 = 0.0608 \).

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Key Concepts

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

Understanding the One-Sample Z-Test
The one-sample z-test is a statistical method used to determine whether there is a significant difference between the mean of a sample and a known population mean. This type of test is particularly useful when you know the population standard deviation, and your sample size is sufficiently large, typically 30 or more entries.

When using the one-sample z-test, you compare your sample mean (\( \bar{x} \)) to a hypothesized population mean (\( \mu \)). The main goal is to assess whether any observed difference is due to random chance or if it indicates a statistically significant disparity. This is done by calculating a \( Z \)-score which shows how far away your sample mean is from the population mean, measured in terms of standard errors. If this score exceeds a critical value from the normal distribution, it leads to rejecting the null hypothesis—the assumption that there is no difference.
Calculating the Standard Error of the Mean
The standard error of the mean (SE Mean) is a critical component in hypothesis testing because it measures how much sample means vary from the true population mean. It acts as a bridge between the sample data and the theoretical concepts of the statistical test.

Calculating the standard error involves dividing the known population standard deviation (\( \sigma \)) by the square root of the sample size (\( n \)). The formula is:\[ SE = \frac{\sigma}{\sqrt{n}} \]This equation reflects that as the sample size increases, its mean becomes a more accurate reflection of the true population mean due to reduced variability.
  • Suppose you have a sample of 16 with a population standard deviation of 1.1—in this case, SE = 0.275.
This calculation helps define the z-value in hypothesis testing, which measures how many SEs the sample mean is away from the population mean.
Navigating P-Value Calculation
The p-value is a fundamental statistical measure that helps determine the significance of your hypothesis test. It represents the probability of obtaining results at least as extreme as the observed data, assuming the null hypothesis is true.

For a one-sample z-test, after calculating the z-score, you can find the corresponding p-value by referencing a Z-table or using statistical software. This value aids in deciding whether to reject the null hypothesis:
  • If the p-value is less than the chosen significance level (commonly 0.05), you reject the null hypothesis as the observed result is unlikely to occur if the null hypothesis were true.
  • In our example, a Z-value close to 1.876 yields a p-value of roughly 0.0304 for a one-tailed test, indicating significant results.
P-values provide a quantitative measure that helps in making data-driven decisions and understanding the impact of your sample on the wider population.
Decoding Confidence Intervals
Confidence intervals offer a range within which the true population parameter is expected to lie with a certain level of confidence, typically 95%. They offer insights into the precision and reliability of your sample estimates.

A 95% confidence interval can be considered as a range that, were you to repeat the experiment 100 times, would include the true mean 95 of those times. For constructing a 95% lower bound in a one-sample z-test, use the formula:\[\bar{x} - Z_{\alpha} \times SE\]where \( \bar{x} \) is the sample mean, \( Z_{\alpha} \) corresponds to the critical value from the Z-distribution (approximately 1.645 for a 95% confidence level in one-tailed tests), and SE is the standard error.
  • For our scenario, this calculation yields a lower bound for the mean of about 14.56.
Confidence intervals provide not only a range estimate but also offer a visual interpretation of the uncertainty related to sample data, aiding in practical decision-making.

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Most popular questions from this chapter

In each of the following situations, state whether it is a correctly stated hypothesis testing problem and why. (a) \(H_{0}: \mu=25, H_{1}: \mu \neq 25\) (b) \(H_{0}: \sigma>10, H_{1}: \sigma=10\) (c) \(H_{0}: \bar{x}=50, H_{1}: \bar{x} \neq 50\) (d) \(H_{0}: p=0.1, H_{1}: p=0.5\) (e) \(H_{0}: s=30, H_{1}: s>30\)

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 we 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.

Medical researchers have developed a new artificial heart constructed primarily of titanium and plastic. The heart will last and operate almost indefinitely once it is implanted in the patient's body, but the battery pack needs to be recharged about every four hours. A random sample of 50 battery packs is selected and subjected to a life test. The average life of these batteries is 4.05 hours. Assume that battery life is normally distributed with standard deviation \(\sigma=0.2\) hour. (a) Is there evidence to support the claim that mean battery life exceeds 4 hours? Use \(\alpha=0.05 .\) (b) What is the \(P\) -value for the test in part (a)? (c) Compute the power of the test if the true mean battery life is 4.5 hours. (d) What sample size would be required to detect a true mean battery life of 4.5 hours if we 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 one-sided confidence bound on the mean life.

Output from a software package is given below: One-Sample Z: Test of \(m u=20\) vs \(>20\) The assumed standard deviation \(=0.75\) $$ \begin{array}{lrrcccc} \text { Variable } & \mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } & \mathrm{Z} & \mathrm{P} \\ \mathrm{x} & 10 & 19.889 & ? & 0.237 & ? & ? \end{array} $$ (a) Fill in the missing items. What conclusions would you draw? (b) Is this a one-sided or a two-sided test? (c) Use the normal table and the above data to construct a \(95 \%\) two-sided \(\mathrm{CI}\) on the mean. (d) What would the \(P\) -value be if the alternative hypothesis is \(H_{1}: \mu \neq 20 ?\)

Consider the test of \(H_{0}: \sigma^{2}=5\) against \(H_{1}: \sigma^{2}<5 .\) Approximate the \(P\) -value for each of the following test statistics. (a) \(x_{0}^{2}=25.2\) and \(n=20\) (b) \(x_{0}^{2}=15.2\) and \(n=12\) (c) \(x_{0}^{2}=4.2\) and \(n=15\)

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