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Chapter 8: Problem 15

Let \(\mu\) denote the true average reaction time to a certain stimulus. For a \(z\) test of \(H_{0}: \mu=5\) versus \(H_{a}: \mu>5\), determine the \(P\)-value for each of the following values of the \(z\) test statistic. \(\begin{array}{lllllll}\text { a. } 1.42 & \text { b. } .90 & \text { c. } 1.96 & \text { d. } 2.48 & \text { e. }-.11\end{array}\)

Short Answer

Expert verified
a. 0.0778, b. 0.1841, c. 0.0250, d. 0.0066, e. 0.5438

Step by step solution

01

Understand the Hypotheses

We are testing the null hypothesis \(H_{0}: \mu = 5\) against the alternative hypothesis \(H_{a}: \mu > 5\). This is a one-tailed test, because we are only interested in whether \(\mu\) is greater than 5.
02

Identify the Test Statistic Type

For each given test statistic value, we will use the standard normal distribution (also known as the \(z\) distribution) to determine the \(P\)-value. The \(P\)-value is the probability that the test statistic is at least as extreme as the observed value, assuming that the null hypothesis is true.
03

Calculate the P-Value for z = 1.42

For a \(z\) test statistic of 1.42, we find the \(P\)-value by calculating the probability that a standard normal variable is greater than 1.42. This is \(P(Z > 1.42)\). Using standard normal distribution tables or a calculator, we find the probability to the left of 1.42 is approximately 0.9222, so the \(P\)-value is \(1 - 0.9222 = 0.0778\).
04

Calculate the P-Value for z = 0.90

For a \(z\) test statistic of 0.90, we calculate \(P(Z > 0.90)\). Using a standard normal distribution table or calculator, the probability to the left of 0.90 is about 0.8159, so the \(P\)-value is \(1 - 0.8159 = 0.1841\).
05

Calculate the P-Value for z = 1.96

For a \(z\) test statistic of 1.96, calculate \(P(Z > 1.96)\). The probability to the left of 1.96 is about 0.9750, so the \(P\)-value is \(1 - 0.9750 = 0.0250\).
06

Calculate the P-Value for z = 2.48

For a \(z\) test statistic of 2.48, calculate \(P(Z > 2.48)\). The cumulative probability to the left of 2.48 is approximately 0.9934, so the \(P\)-value is \(1 - 0.9934 = 0.0066\).
07

Calculate the P-Value for z = -0.11

For a \(z\) test statistic of -0.11, calculate \(P(Z > -0.11)\). Since the normal distribution is symmetric, \(P(Z > -0.11)\) is \(1 - 0.4562 = 0.5438\), where 0.4562 is the probability to the left of -0.11.

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Key Concepts

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

z test
A z test is a statistical method used to determine whether there is a significant difference between the means of two groups or a sample mean and a known population mean. In the context of the original exercise, the z test is used to test a hypothesis about the average reaction time to a stimulus.

Here's a simple breakdown of how a z test works:
  • Firstly, it begins with the formulation of the null hypothesis (\(H_0\): \(\mu=5\)) and the alternative hypothesis ( \(H_a\): \(\mu>5\)). The null hypothesis acts as a statement of no effect or no difference, while the alternative suggests a potential effect.
  • We then calculate the z test statistic, which measures how many standard deviations our sample mean is from the known population mean.
  • A z test is particularly useful when the sample size is large, typically greater than 30, or when the population standard deviation is known.
Think of the z test as a way of assessing whether any observed difference in data is likely due to random chance or because there truly is a difference.
standard normal distribution
The standard normal distribution, also referred to as the z distribution, is a special kind of normal distribution. It has a mean (\(\mu\)) of 0 and a standard deviation (\(\sigma\)) of 1. The values we calculate for the z test (known as z scores) are compared against this distribution.

Key characteristics of the standard normal distribution include:
  • It is symmetrical around the mean, meaning it looks like a classic bell curve.
  • Since it is standardized, any normal distribution can be transformed into it by using the z-score formula: \(z = \frac{(X - \mu)}{\sigma}\), where \(X\) is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
  • Areas under the curve represent probabilities, showing the likelihood of a z score falling within a specified range.
For our exercise, we refer to tables or calculators of the standard normal distribution to find probabilities for given z scores. These values help us determine how likely it is to observe a test statistic, assuming the null hypothesis is true.
p-value calculation
The p-value is a key component in hypothesis testing used to determine the significance of the results. It tells us the probability of observing a test statistic as extreme as, or more extreme than, the one calculated, assuming the null hypothesis is true.

Steps for calculating the p-value include:
  • Compute the z test statistic from your sample data as mentioned before.
  • Use the standard normal distribution to find the probability that the calculated z score or a more extreme value occurs. This is done by looking up the z score in a z table or using a calculator.
  • For one-tailed tests like in our exercise, you subtract the cumulative probability found from 1 to get the p-value. Thus, \(P(Z > z)\) is found as \(1 - P(Z < z)\).
A smaller p-value suggests that the observed data is unlikely under the null hypothesis, leading to stronger evidence against it. Conversely, a larger p-value indicates that the data could easily occur under the null hypothesis, implying less evidence against it.

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Most popular questions from this chapter

The recommended daily dietary allowance for zinc among males older than age 50 years is \(15 \mathrm{mg} /\) day. The article "Nutrient Intakes and Dietary Patterns of Older Americans: A National Study" \((J .\) of Gerontology, 1992: M145-150) reports the following summary data on intake for a sample of males age 65-74 years: \(n=115, \bar{x}=11.3\), and \(s=6.43\). Does this data indicate that average daily zinc intake in the population of all males ages \(65-74\) falls below the recommended allowance?

Many older homes have electrical systems that use fuses rather than circuit breakers. A manufacturer of 40 -amp fuses wants to make sure that the mean amperage at which its fuses burn out is in fact 40 . If the mean amperage is lower than 40 , customers will complain because the fuses require replacement too often. If the mean amperage is higher than 40 , the manufacturer might be liable for damage to an electrical system due to fuse malfunction. To verify the amperage of the fuses, a sample of fuses is to be selected and inspected. If a hypothesis test were to be performed on the resulting data, what null and alternative hypotheses would be of interest to the manufacturer? Describe type I and type II errors in the context of this problem situation.

The following observations are on stopping distance (ft) of a particular truck at \(20 \mathrm{mph}\) under specified experimental conditions ("Experimental Measurement of the Stopping Performance of a 'Tractor-Semitrailer from Multiple Speeds" NHTSA, DOT HS 811488 , June 2011): \(\begin{array}{llllll}32.1 & 30.6 & 31.4 & 30.4 & 31.0 & 31.9\end{array}\) The cited report states that under these conditions, the maximum allowable stopping distance is 30 . A normal probability plot validates the assumption that stopping distance is normally distributed. a. Does the data suggest that true average stopping distance exceeds this maximum value? Test the appropriate hypotheses using \(\alpha=.01\). b. Determine the probability of a type II error when \(\alpha=\) \(.01, \sigma=.65\), and the actual value of \(\mu\) is 31 . Repeat this for \(\mu=32\) (use either statistical software or Table A.17). c. Repeat (b) using \(\sigma=.80\) and compare to the results of \((b)\). d. What sample size would be necessary to have \(\alpha=\) \(.01\) and \(\beta=.10\) when \(\mu=31\) and \(\sigma=.65 ?\)

Before agreeing to purchase a large order of polyethylene sheaths for a particular type of high-pressure oilfilled submarine power cable, a company wants to see conclusive evidence that the true standard deviation of sheath thickness is less than \(.05 \mathrm{~mm}\). What hypotheses should be tested, and why? In this context, what are the type I and type II errors?

For each of the following assertions, state whether it is a legitimate statistical hypothesis and why: a. \(H: \sigma>100\) b. \(H: \tilde{x}=45\) c. \(H: s \leq .20\) d. \(H: \sigma_{1} / \sigma_{2}<1\) e. \(H: \bar{X}-\bar{Y}=5\) f. \(H: \lambda \leq .01\), where \(\lambda\) is the parameter of an exponential distribution used to model component lifetime
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