/*! 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} Q4E Pairs of P-values and significan... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Q4E (page 325)

Pairs of P-values and significance levels, 伪, are given.

For each pair, state whether the observed P-value would lead to rejection of H0 at the given significance level.

a.P颅-value = .084, 伪= .05

b.P颅-value = .003, 伪= .001

c.P-颅value = .498, 伪= .05

d.P-颅value = .084, 伪= .10

e.P-颅value = .039, 伪= .01

f.P-颅value = .218, 伪 = .10

Short Answer

Expert verified

a)Fail to reject the null hypothesis\({H_0}\).

b)Fail to reject the null hypothesis\({H_0}\).

c)Fail to reject the null hypothesis\({H_0}\).

d)Reject the null hypothesis\({H_0}\).

e)Fail to reject the null hypothesis\({H_0}\).

f)Fail to reject the null hypothesis \({H_0}\).

Step by step solution

01

Step 1:Null hypothesis.

The null hypothesis, denoted by H0, is the claim that is initially assumed to be true (the 鈥減rior belief鈥� claim). The alternative hypothesis, denoted by Ha, is the assertion that is contradictory to H0.

The null hypothesis will be rejected in favour of the alternative hypothesis only if sample evidence suggests that H0 is false. If the sample does not strongly contradict H0, we will continue to believe in the plausibility of the null hypothesis. The two possible conclusions from a hypothesis-testing analysis are then reject H0 or fail to reject H0.

02

Step 2:Solution for part a).

Given that,

\(\begin{array}{l}P = 0.084\\\alpha = 0.05\end{array}\)

If the \(P\)-value is less than the significance level \(\alpha \), then reject the null hypothesis \({H_0}\), else we fail to reject the null hypothesis \({H_0}\).

\(P = 0.084 > 0.05\)

In the given \(P > \alpha \), so, it fails to rejects the \({H_0}\).

Fail to reject the null hypothesis \({H_0}\).

03

Step 3:Solution for part b).

Given that,

\(\begin{array}{l}P = 0.003\\\alpha = 0.001\end{array}\)

If the \(P\)-value is less than the significance level \(\alpha \), then reject the null hypothesis \({H_0}\), else we fail to reject the null hypothesis \({H_0}\).

\(P = 0.003 > 0.001\)

In the given \(P > \alpha \), so, it fails to rejects the \({H_0}\).

Fail to reject the null hypothesis \({H_0}\).

04

Step 4:Solution for part c).

Given that,

\(\begin{array}{l}P = 0.498\\\alpha = 0.05\end{array}\)

If the \(P\)-value is less than the significance level \(\alpha \), then reject the null hypothesis \({H_0}\), else we fail to reject the null hypothesis \({H_0}\).

\(P = 0.498 > 0.05\)

In the given \(P > \alpha \), so, it fails to rejects the \({H_0}\).

Fail to reject the null hypothesis \({H_0}\).

05

Step 5:Solution for part d).

Given that,

\(\begin{array}{l}P = 0.084\\\alpha = 0.10\end{array}\)

If the \(P\)-value is less than the significance level \(\alpha \), then reject the null hypothesis \({H_0}\), else we fail to reject the null hypothesis \({H_0}\).

\(P = 0.084 < 0.10\)

In the given \(P < \alpha \), so, it rejects the \({H_0}\).

06

Step 6:Solution for part e).

Given that,

\(\begin{array}{l}P = 0.039\\\alpha = 0.01\end{array}\)

If the \(P\)-value is less than the significance level \(\alpha \), then reject the null hypothesis \({H_0}\), else we fail to reject the null hypothesis \({H_0}\).

\(P = 0.039 > 0.01\)

In the given \(P > \alpha \), so, it fails to rejects the \({H_0}\).

Fail to reject the null hypothesis \({H_0}\).

07

Step 7:Solution for part f).

Given that,

\(\begin{array}{l}P = 0.218\\\alpha = 0.10\end{array}\)

If the \(P\)-value is less than the significance level \(\alpha \), then reject the null hypothesis \({H_0}\), else we fail to reject the null hypothesis \({H_0}\).

\(P = 0.218 > 0.10\)

In the given \(P > \alpha \), so, it fails to rejects the \({H_0}\).

Fail to reject the null hypothesis \({H_0}\).

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!

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

The following observations are on stopping distance (ft) of a particular truck at \(20mph\) under specified experimental conditions (鈥淓xperimental Measurement of the Stopping Performance of a Tractor-Semitrailer from Multiple Speeds,鈥� NHTSA, DOT HS 811 488, June 2011):

\(32.1 30.6 31.4 30.4 31.0 31.9\)

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 a 5 .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 \(\sigma = .65\)?

A random sample of \(150\) recent donations at a certain blood bank reveals that \(82\) were type A blood. Does this suggest that the actual percentage of type A donations differs from \(40\% \), the percentage of the population having type A blood? Carry out a test of the appropriate hypotheses using a significance level of \(.01\). Would your conclusion have been different if a significance level of \(.05\) had been used?

In Problems \(1 - 8\) state the order of the given ordinary differential equation. Determine whether the equation is linear or nonlinear by matching it with \((6)\).

\((sin\theta )y''' - (cos\theta )y' = 2\)

For each of the following assertions, state whether it is a legitimate statistical hypothesis and why:

\(\begin{array}{l}{\rm{a}}{\rm{. H: \sigma > 100}}\\{\rm{c}}{\rm{. H: s }} \le {\rm{.20}}\\{\rm{e}}{\rm{. H:}}\overline {{\rm{ X}}} {\rm{ - }}\overline {\rm{Y}} {\rm{ = 5}}\end{array}\) \(\begin{array}{l}{\rm{b}}{\rm{. H: }}\widetilde {\rm{x}}{\rm{ = 45}}\\{\rm{d}}{\rm{. H: }}{{\rm{\sigma }}_{\rm{1}}}{\rm{/}}{{\rm{\sigma }}_{\rm{2}}}{\rm{ < 1}}\end{array}\)

\({\rm{f}}{\rm{. H: \lambda }} \le {\rm{.01}}\), where \({\rm{\lambda }}\) is the parameter of an exponential distribution used to model component lifetime

In Problems \(1 - 8\) state the order of the given ordinary differential equation. Determine whether the equation is linear or nonlinear by matching it with \((6)\).

\(x\frac{{{d^3}y}}{{d{x^3}}} - {\left( {\frac{{dy}}{{dx}}} \right)^4} + y = 0\)
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

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.