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Chapter 9: Problem 37

For the hypothesis test \(H_{0}: \mu=5\) against \(H_{1}: \mu<5\) and variance known, calculate the \(P\) -value for each of the following test statistics. (a) \(z_{0}=2.05\) (b) \(z_{0}=-1.84\) (c) \(z_{0}=0.4\)

Short Answer

Expert verified
(a) P-value = 0.0202, (b) P-value = 0.0329, (c) P-value = 0.6554.

Step by step solution

01

Understanding the Hypothesis Test

We have a null hypothesis \(H_0: \mu = 5\) and an alternative hypothesis \(H_1: \mu < 5\). We are testing if the true mean \(\mu\) is less than 5. We are given the test statistic values \(z_0\) for three different cases.
02

Determine P-value for Test Statistic (a)

For \(z_0 = 2.05\), we check the standard normal distribution (Z-distribution) to find the P-value. Since \(z_0\) is greater than zero, it indicates the test statistic is on the right side of the mean. However, since our \(H_1\) is \(\mu < 5\), we consider the left tail. The P-value is \(1 - P(Z < 2.05)\). Calculating this, \(P(Z < 2.05) \approx 0.9798\), so the P-value is approximately \(1 - 0.9798 = 0.0202\).
03

Determine P-value for Test Statistic (b)

For \(z_0 = -1.84\), we find the P-value using the Z-distribution by looking for the probability to the left of \(z_0\). Since it is a left-tailed test and \(z_0\) is negative, the P-value is \(P(Z < -1.84)\). Calculating this, \(P(Z < -1.84) \approx 0.0329\).
04

Determine P-value for Test Statistic (c)

For \(z_0 = 0.4\), look up the P-value for a standard normal distribution. Since both the test and the hypothesis are left-tailed, the P-value is \(P(Z < 0.4)\). Calculating this, \(P(Z < 0.4) \approx 0.6554\).
05

Conclusion for Each Hypothesis Test

We've found the P-values: for (a) it is 0.0202, for (b) it is 0.0329, and for (c) it is 0.6554. These values tell us the probability of observing the test statistic or something more extreme under the null hypothesis, \(H_0\).

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Key Concepts

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

P-value calculation
The P-value is a crucial concept in hypothesis testing in statistics. It represents the probability of obtaining test results at least as extreme as the observed results, under the assumption that the null hypothesis is correct.
Calculating the P-value involves using the test statistic from your experiment or study. Here's a simple breakdown:
  • Identify the test statistic (e.g.,\( z_0 \)) used to measure the deviation from the null hypothesis.
  • Utilize the standard normal distribution (Z-distribution) to determine how extreme the test statistic is.
  • Compute the probability corresponding to the deviation on the normal distribution curve. This helps determine whether the result is significant.
Understanding these probabilities allows researchers to decide if they should reject the null hypothesis in favor of the alternative hypothesis. Smaller P-values indicate stronger evidence against the null hypothesis.
Standard Normal Distribution
The standard normal distribution, often referred to as the Z-distribution, is a special normal distribution with a mean of 0 and a standard deviation of 1. It is symmetric around the mean, and it is one of the most important probability distributions in statistics.
  • The standard normal distribution is used to convert normal distributions into a standardized form, making it easier to calculate probabilities and critical values.
  • The Z-score, or standard score, helps in determining the distance of a data point from the mean. It is given by the formula: \( Z = \frac{X - \mu}{\sigma} \) where \( X \) is a data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
  • Z-tables or standard normal tables help in finding cumulative probabilities associated with Z-scores, which are essential for P-value calculations.
By using standardization, you can compare two different datasets effectively or conduct hypothesis tests efficiently.
Null and Alternative Hypotheses
In hypothesis testing, the null hypothesis \((H_0)\) and the alternative hypothesis \((H_1)\) form the foundation of the method.
  • The null hypothesis \((H_0)\) represents a statement of no effect or no difference. It is the hypothesis that an experiment aims to test.
  • The alternative hypothesis \((H_1)\) indicates the presence of an effect or a difference. It contrasts with the null hypothesis.
  • Three main types of alternative hypotheses include:
    1. Two-tailed (\( H_1: \mu eq \mu_0 \))
    2. Left-tailed (\( H_1: \mu < \mu_0 \))
    3. Right-tailed (\( H_1: \mu > \mu_0 \))
Hypothesis testing determines if there is enough statistical evidence to reject the null hypothesis in favor of the alternative. This process aids in making informed conclusions based on sample data.

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Most popular questions from this chapter

Consider the hypothesis test of \(H_{0}: \sigma^{2}=10\) against \(H_{1}: \sigma^{2}>10 .\) 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 \(n=15\)

A hypothesis will be used to test that a population mean equals 10 against the alternative that the population mean is more than 10 with known variance \(\sigma\). What is the critical value for the test statistic \(Z_{0}\) for the following significance levels? (a) 0.01 (b) 0.05 (c) 0.10

A semiconductor manufacturer collects data from a new tool and conducts a hypothesis test with the null hypothesis that a critical dimension mean width equals \(100 \mathrm{nm}\). The conclusion is to not reject the null hypothesis. Does this result provide strong evidence that the critical dimension mean equals \(100 \mathrm{nm}\) ? Explain.

An article in Biological Trace Element Research \(\left[{ }^{\circ \cdot}\right.\) Interaction of Dietary Calcium, Manganese, and Manganese Source (Mn Oxide or Mn Methionine Complex) or Chick Performance and Manganese Utilization" (1991, Vol. 29(3), pp. 217-228)] showed the following results of tissue assay for liver manganese (ppm) in chicks fed high Ca diets. $$\begin{array}{llllll}6.02 & 6.08 & 7.11 & 5.73 & 5.32 & 7.10 \\\5.29 & 5.84 & 6.03 & 5.99 & 4.53 & 6.81\end{array}$$ (a) Test the hypothesis \(H_{0}: \sigma^{2}=0.6\) versus \(H_{1}: \sigma^{2} \neq 0.6\) using\(\alpha=0.01\) (b) What is the \(P\) -value for this test? (c) Discuss how part (a) could be answered by constructing a \(99 \%\) two-sided confidence interval for \(\sigma .\)

Exercise \(6-40\) presented data on the concentration of suspended solids in lake water. (a) Test the hypothesis \(H_{0}: \mu=55\) versus \(H_{1}: \mu \neq 55 ;\) use \(\alpha=0.05 .\) Find the \(P\) -value. (b) Check the normality assumption. (c) Compute the power of the test if the true mean concentration is as low as 50 . (d) What sample size would be required to detect a true mean concentration as low as 50 if you wanted the power of the test to be at least \(0.9 ?\)
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