/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 In an experiment designed to mea... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 31

In an experiment designed to measure the time necessary for an inspector's eyes to become used to the reduced amount of light necessary for penetrant inspection, the sample average time for \(n=9\) inspectors was \(6.32 \mathrm{sec}\) and the sample standard deviation was \(1.65 \mathrm{sec}\). It has previously been assumed that the average adaptation time was at least \(7 \mathrm{sec}\). Assuming adaptation time to be normally distributed, does the data contradict prior belief? Use the \(t\) test with \(\alpha=.1\).

Short Answer

Expert verified
No, the data does not significantly contradict the prior belief at \( \alpha = 0.1 \).

Step by step solution

01

State the Hypotheses

First, we need to set up the null and alternative hypotheses. The null hypothesis, \( H_0 \), states that the average adaptation time is at least 7 seconds. The alternative hypothesis, \( H_a \), states that the average adaptation time is less than 7 seconds. \[H_0: \mu \geq 7 \H_a: \mu < 7\]
02

Determine the Test Statistic

Since the sample size \( n = 9 \) is small and the population standard deviation is unknown, we use the \( t \)-test. The test statistic \( t \) is calculated using the following formula:\[t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}\]where \( \bar{x} = 6.32 \), \( \mu_0 = 7 \), \( s = 1.65 \), and \( n = 9 \). Substituting these values, we get:\[t = \frac{6.32 - 7}{\frac{1.65}{\sqrt{9}}} = \frac{-0.68}{0.55} \approx -1.236\]
03

Determine the Critical Value

Determine the critical \( t \)-value from the \( t \)-distribution table for a one-tailed test with \( \alpha = 0.1 \) and \( df = n - 1 = 8 \). The critical \( t \)-value at \( \alpha = 0.1 \) for 8 degrees of freedom is approximately \( -1.397 \).
04

Make the Decision

Compare the calculated \( t \)-statistic to the critical \( t \)-value. The calculated \( t \)-statistic is \( -1.236 \) and the critical \( t \)-value is \( -1.397 \). Since \( -1.236 \) is greater than \( -1.397 \), we fail to reject the null hypothesis.
05

Conclusion

There is not enough evidence at the \( \alpha = 0.1 \) level to conclude that the average adaptation time is less than 7 seconds. The data does not significantly contradict the prior belief.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Hypothesis Testing
Hypothesis testing is a statistical method used to make inferences about a population based on a sample. It begins with formulating two hypotheses: the null hypothesis
  • Null Hypothesis (\(H_0\)): A statement suggesting there is no effect or no difference. In this case, it claims that the average adaptation time is at least 7 seconds.
  • Alternative Hypothesis (\(H_a\)): A statement that contradicts the null hypothesis. Here, it suggests the average adaptation time is less than 7 seconds.
Hypothesis testing then involves collecting sample data and using it to determine whether there is enough evidence to reject the null hypothesis. This decision is based on the calculated test statistic and comparing it to a critical value, which is determined by the significance level \(\alpha\), in this case, 0.1. This structured approach helps researchers discern if observed data significantly contradict a believed standard or assumption.
Sample Statistics
Sample statistics provide valuable information derived from the data collected from a sample. They summarize attributes of the data and are crucial in the context of hypothesis testing.
  • Sample Mean (\(\bar{x}\)): The average of all observations in the sample. Here, it is 6.32 seconds.
  • Sample Standard Deviation (\(s\)): Indicates how spread out the sample data is around the mean. In the problem, it is 1.65 seconds.
  • Sample Size (\(n\)): The number of observations in the sample. For this case, it is 9.
Sample statistics are used to compute the test statistic, a value that helps in evaluating if the sample data is significantly different from the population average, given the null hypothesis.
Critical Value
The critical value in hypothesis testing is the threshold that the calculated test statistic must exceed in order to reject the null hypothesis. It depends on:
  • Significance Level (\(\alpha\)): The probability of rejecting the null hypothesis when it is true. Here, \(\alpha = 0.1\)
  • Degrees of Freedom (df): For a t-test, it is calculated as \(n-1\). In this problem, \(df = 8\)
By consulting a t-distribution table, we find that the critical value for a one-tailed test at \(\alpha = 0.1\) with 8 degrees of freedom is approximately \(-1.397\). If the test statistic is less than (more negative than) this critical value in the context of a one-tailed test, the null hypothesis would be rejected. However, in this case, the test statistic of \(-1.236\) did not surpass the critical threshold, leading to a failure in rejecting \(H_0\).

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A sample of 50 lenses used in eyeglasses yields a sample mean thickness of \(3.05 \mathrm{~mm}\) and a sample standard deviation of \(.34 \mathrm{~mm}\). The desired true average thickness of such lenses is \(3.20 \mathrm{~mm}\). Does the data strongly suggest that the true average thickness of such lenses is something other than what is desired? Test using \(\alpha=.05\).

To determine whether the pipe welds in a nuclear power plant meet specifications, a random sample of welds is selected, and tests are conducted on each weld in the sample. Weld strength is measured as the force required to break the weld. Suppose the specifications state that mean strength of welds should exceed \(100 \mathrm{lb} / \mathrm{in}^{2}\); the inspection team decides to test \(H_{0}: \mu=100\) versus \(H_{\mathrm{a}}: \mu>100\). Explain why it might be preferable to use this \(H_{\mathrm{a}}\) rather than \(\mu<100\).

When the population distribution is normal and \(n\) is large, the sample standard deviation \(S\) has approximately a normal distribution with \(E(S) \approx \sigma\) and \(V(S) \approx \sigma^{2} /(2 n)\). We already know that in this case, for any \(n, \bar{X}\) is normal with \(E(\bar{X})=\mu\) and \(V(\bar{X})=\sigma^{2} / n\) a. Assuming that the underlying distribution is normal, what is an approximately unbiased estimator of the 99 th percentile \(\theta=\mu+2.33 \sigma\) ? b. When the \(X_{i}\) s are normal, it can be shown that \(\bar{X}\) and \(S\) are independent rv's (one measures location whereas the other measures spread). Use this to compute \(V(\hat{\theta})\) and \(\sigma_{\hat{\theta}}\) for the estimator \(\hat{\theta}\) of part (a). What is the estimated standard error \(\hat{\sigma}_{\hat{\theta}}\) ?

The article "Orchard Floor Management Utilizing SoilApplied Coal Dust for Frost Protection" (Agri. and Forest Meteorology, 1988: 71-82) reports the following values for soil heat flux of eight plots covered with coal dust. \(\begin{array}{llllllll}34.7 & 35.4 & 34.7 & 37.7 & 32.5 & 28.0 & 18.4 & 24.9\end{array}\) The mean soil heat flux for plots covered only with grass is 29.0. Assuming that the heat-flux distribution is approximately normal, does the data suggest that the coal dust is effective in increasing the mean heat flux over that for grass? Test the appropriate hypotheses using \(\alpha=.05\).

The incidence of a certain type of chromosome defect in the U.S. adult male population is believed to be 1 in 75 . A random sample of 800 individuals in U.S. penal institutions reveals 16 who have such defects. Can it be concluded that the incidence rate of this defect among prisoners differs from the presumed rate for the entire adult male population? a. State and test the relevant hypotheses using \(\alpha=.05\). What type of error might you have made in reaching a conclusion? b. What \(P\)-value is associated with this test? Based on this \(P\)-value, could \(H_{0}\) be rejected at significance level .20?
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.