/*! 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} Q9E Question: Independent random sam... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Q9E (page 452)

Question: Independent random samples n1 =233 and n2=312 are selected from two populations and used to test the hypothesis Ha:(渭1-渭)2=0against the alternative Ha:(渭1-渭)2鈮�0

.a. The two-tailed p-value of the test is 0.1150 . Interpret this result.b. If the alternative hypothesis had been Ha:(渭1-渭)2<0 , how would the p-value change? Interpret the p-value for this one-tailed test.

Short Answer

Expert verified

Answer

Random sampling guarantees that the findings received from your sample are close to those received if the full sample was surveyed.

Step by step solution

01

(a) Interpret the result for the given two-tailed p-value of the test

The null hypothesis is Ha:渭1-渭2=0

If we take the significance level to be 0.05 and the p-value is 0.1150.

As the p-value is more than the significance level, so the null hypothesis will not be rejected.

If we take the significance level to be 0.01 , and the p-value is 0.1150 .

As the p-value is more than the significance level, so the null hypothesis will not be rejected.

02

(b) Interpret the p-valueHa:(μ1-μ)2<0  .

If the alternative hypothesis has changed from Ha:渭1-渭2鈮�0toHa:渭1-渭2<0 , it implies that the p-value should be halved as now it is a one-tailed test instead of a two-tailed test.

So, the p-value will be 0.0575 .

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

Given that xis a hypergeometric random variable, computep(x)for each of the following cases:

a. N= 8, n= 5, r= 3, x= 2

b. N= 6, n= 2, r= 2, x= 2

c. N= 5, n= 4, r= 4, x= 3

The gender diversity of a large corporation鈥檚 board of directors was studied in Accounting & Finance (December 2015). In particular, the researchers wanted to know whether firms with a nominating committee would appoint more female directors than firms without a nominating committee. One of the key variables measured at each corporation was the percentage of female board directors. In a sample of 491firms with a nominating committee, the mean percentage was 7.5%; in an independent sample of 501firms without a nominating committee, the mean percentage was role="math" localid="1652702402701" 4.3% .

a. To answer the research question, the researchers compared the mean percentage of female board directors at firms with a nominating committee with the corresponding percentage at firms without a nominating committee using an independent samples test. Set up the null and alternative hypotheses for this test.

b. The test statistic was reported as z=5.1 with a corresponding p-value of 0.0001. Interpret this result if 伪=0.05.

c. Do the population percentages for each type of firm need to be normally distributed for the inference, part b, to be valid? Why or why not?

d. To assess the practical significance of the test, part b, construct a 95% confidence interval for the difference between the true mean percentages at firms with and without a nominating committee. Interpret the result.

Comparing taste-test rating protocols. Taste-testers of new food products are presented with several competing food samples and asked to rate the taste of each on a 9-point scale (where1="dislike extremely" and9="like extremely"). In the Journal of Sensory Studies (June 2014), food scientists compared two different taste-testing protocols. The sequential monadic (SM) method presented the samples one-at-a-time to the taster in a random order, while the rank rating (RR) method presented the samples to the taster all at once, side-by-side. Consider the following experiment (similar to the one conducted in the journal): 50 consumers of apricot jelly were asked to taste test five different varieties. Half the testers used the SM protocol and half used the RR protocol during testing. In a second experiment, 50 consumers of cheese were asked to taste-test four different varieties. Again, half the testers used the SM protocol and half used the RR protocol during testing. For each product (apricot jelly and cheese), the mean taste scores of the two protocols (SM and RR) were compared. The results are shown in the accompanying tables.

a. Consider the five varieties of apricot jelly. Identify the varieties for which you can conclude that "the mean taste scores of the two protocols (SM and RR) differ significantly at伪=.05."

b. Consider the four varieties of cheese. Identify the varieties for which you can conclude that "the mean taste scores of the two protocols (SM and RR) differ significantly at伪=.05."

c. Explain why the taste-test scores do not need to be normally distributed for the inferences, parts a and b, to be valid.

To use the t-statistic to test for a difference between the means of two populations, what assumptions must be made about the two populations? About the two samples?

Tomato as a taste modifier. Miraculin鈥攁 protein naturally produced in a rare tropical fruit鈥攈as the potential to be an alternative low-calorie sweetener. In Plant Science (May2010), a group of Japanese environmental scientists investigated the ability of a hybrid tomato plant to produce miraculin. For a particular generation of the tomato plant, the amount x of miraculin produced (measured in micrograms per gram of fresh weight) had a mean of 105.3 and a standard deviation of 8.0. Assume that x is normally distributed.

a. FindP(x>120).

b. FindP(100<x<110).

c. Find the value a for whichP(x<a)=0.25.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Geometry

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.