Beginning Statistics, 2nd Edition
Beginning Statistics, 2nd Edition
2nd Edition
ISBN: 9781932628678
Author: Carolyn Warren; Kimberly Denley; Emily Atchley
Publisher: Hawkes Learning Systems
Question
Book Icon
Chapter 11.1, Problem 19E
To determine

(a)

To find:

The null and alternative hypotheses.

Expert Solution
Check Mark

Answer to Problem 19E

Solution:

The null hypothesis is H0:μ1μ20 and the alternative hypothesis is H1:μ1μ2>0.

Explanation of Solution

Given information:

Lauren and Keri live in different states and disagree about who has higher electric bills. To settle their disagreement, the girls decided to sample electric bills in their area for the month of June and perform a hypothesis test that the mean monthly residential electric bill is higher for Lauren’s state than for Keri’s state. Is there evidence at the 0.01 Level to say that the mean monthly residential electric bill is higher for Lauren’s state than for Keri’s state.

The null hypothesis is a statement of no difference, that there is no significant difference between the two phenomena. It is considered to be true until it is nullified by statistical evidence for an alternative hypothesis.

An alternative hypothesis is a contradicting statement to the null hypothesis and states a significant difference between the two phenomena It is accepted when the null hypothesis is false.

Calculation:

From the given information:

Let μ1 and μ2 be the mean electric bill of the company in Lauren’s and Keri’s states respectively.

From the given information, the claim is to test that the residential electric bill is higher for Lauren’s state than for Keri’s state, i.e.

μ1>μ2μ1μ2>0

Therefore the null hypothesis is H0:μ1μ20 and the alternative hypothesis, H1:μ1μ2>0.

To determine

(b)

To find:

Which distribution to be used for test statistic, and the level of significance.

Expert Solution
Check Mark

Answer to Problem 19E

Solution:

The distribution used is standard normal variate (one-tailed z test), and the level of significance is 1% or 0.01.

Explanation of Solution

Since both the sample sizes are greater than 30, normal distribution will be followed and the standard normal variate z-test will be applied.

According to the null hypothesis, one-tailed test is to be applied.

From the given information, the level of significance is 1% or 0.01.

To determine

(c)

To find:

The test statistic

Expert Solution
Check Mark

Answer to Problem 19E

Solution:

The test statistic is 0.66

Explanation of Solution

Given information:

For Lauren’s state:

The sample size is 35, the mean monthly electric bill is $104.53 and the population standard deviation is $17.81.

For Keri’s state:

The sample size is 51, the mean monthly electric bill is $101.48 and the population standard deviation is $25.30.

Test statistics is a random variable which is calculated from the sample data and used in hypothesis testing.

Test statistic calculates the degrees of acceptance between sample data and null hypothesis.

Formula used:

The test statistic for a hypothesis test for two population means is:

z=(x¯1x¯2)(μ1μ2)σ12n1+σ22n2

Where: x¯1 and x¯2 are the two sample means,

μ1μ2 is the hypothetical value of the difference between the two population means,

σ1 and σ2 are the two population standard deviations,

and n1 and n2 are the two sample sizes.

Calculation:

The test statistic for a given hypothesis is,

z=(x¯1x¯2)(μ1μ2)σ12n1+σ22n2

Substitute 104.53 for x¯1, 17.81 for σ1, 35 for n1, 101.48 for x¯2, 25.30 for σ2 and 51 for n2 in above equation.

z=(104.53101.48)(μ1μ2)(17.81)235+(25.30)251

It is given that the null hypothesis for the given proportion is H0:μ1μ20. Substitute 0 for μ1μ2.

z=3.0509.0627+12.5508=3.054.6490=0.65610.66

Thus, the test statistic is 0.66.

To determine

(d)

To find:

The conclusion by comparing the p-value to the level of significance.

Expert Solution
Check Mark

Answer to Problem 19E

Solution:

The null hypothesis is accepted and it shows that there is no significant difference in the electric bills of the two companies.

There is insufficient evidence to say that the electric bill of the company in Lauren’s state is higher than by Keri’s state.

Explanation of Solution

p-value or probability value in a given statistical hypothesis is used to determine the significance of the results. A small p-value (mainly 0.05) indicates evidence against the null hypothesis.

Formula used:

The p- value is given by P(zZ), where Z is the value of the test statistic. p-value is equivalent to the area under the standard normal curve to the right of z=Z.

Calculation:

This is a right-tailed test, so p-value = P (z0.66)

p-value is equivalent to the area under the standard normal curve to the right of z=0.66

p-value =0.2546 at 0.01 level of significance.

Conclusion:

Test statistic is 0.66 and the p-value is 0.2546.

Since the p-value is >0.01, the null hypothesis is accepted.

The null hypothesis is accepted and it shows that there is no significant difference in the electric bills of the two companies.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
SCIE 211 Lab 3: Graphing and DataWorksheetPre-lab Questions:1. When should you use each of the following types of graphs? Fill answers in the table below.Type of Graph Used to showLine graphScatter plotBar graphHistogramPie Chart2. Several ways in which we can be fooled or misled by a graph were identified in the Lab 3Introduction. Find two examples of misleading graphs on the Internet and paste them below. Besure to identify why each graph is misleading.                 Data Charts:Circumference vs. Diameter for circular objectsDiameter Can 1 (cm) Can 2 (cm) Can 3 (cm)Trial 1Trial 2Trial 3MeanCircumference Can 1 (cm) Can 2 (cm) Can 3 (cm)Trial 1Trial 2Trial 3MeanScatter Plot Graph – Circumference Vs. DiameterIdentify 2 points of the Trendline.Y1 = ________ Y2 = _________X1 = ________ X2 = _________Calculate the Slope of the Trendline =                 Post-lab Questions:1. Answer the questions below. You will need to use the following equation to answer…
The U.S. Bureau of Labor Statistics reports that 11.3% of U.S. workers belong to unions (BLS website, January 2014). Suppose a sample of 400 U.S. workers is collected in 2014 to determine whether union efforts to organize have increased union membership. a.  Formulate the hypotheses that can be used to determine whether union membership increased in 2014.H 0: p H a: p  b.  If the sample results show that 52 of the workers belonged to unions, what is the p-value for your hypothesis test (to 4 decimals)?
A company manages an electronic equipment store and has ordered 200200 LCD TVs for a special sale. The list price for each TV is ​$200200 with a trade discount series of 6 divided by 10 divided by 2.6/10/2. Find the net price of the order by using the net decimal equivalent.

Chapter 11 Solutions

Beginning Statistics, 2nd Edition

Ch. 11.1 - Prob. 11ECh. 11.1 - Prob. 12ECh. 11.1 - Prob. 13ECh. 11.1 - Prob. 14ECh. 11.1 - Prob. 15ECh. 11.1 - Prob. 16ECh. 11.1 - Prob. 17ECh. 11.1 - Prob. 18ECh. 11.1 - Prob. 19ECh. 11.1 - Prob. 20ECh. 11.1 - Prob. 21ECh. 11.2 - Prob. 1ECh. 11.2 - Prob. 2ECh. 11.2 - Prob. 3ECh. 11.2 - Prob. 4ECh. 11.2 - Prob. 5ECh. 11.2 - Prob. 6ECh. 11.2 - Prob. 7ECh. 11.2 - Prob. 8ECh. 11.2 - Prob. 9ECh. 11.2 - Prob. 10ECh. 11.2 - Prob. 11ECh. 11.2 - Prob. 12ECh. 11.2 - Prob. 13ECh. 11.2 - Prob. 14ECh. 11.2 - Prob. 15ECh. 11.2 - Prob. 16ECh. 11.2 - Prob. 17ECh. 11.3 - Prob. 1ECh. 11.3 - Prob. 2ECh. 11.3 - Prob. 3ECh. 11.3 - Prob. 4ECh. 11.3 - Prob. 5ECh. 11.3 - Prob. 6ECh. 11.3 - Prob. 7ECh. 11.3 - Prob. 8ECh. 11.3 - Prob. 9ECh. 11.3 - Prob. 10ECh. 11.4 - Prob. 1ECh. 11.4 - Prob. 2ECh. 11.4 - Prob. 3ECh. 11.4 - Prob. 4ECh. 11.4 - Prob. 5ECh. 11.4 - Prob. 6ECh. 11.4 - Prob. 7ECh. 11.4 - Prob. 8ECh. 11.4 - Prob. 9ECh. 11.4 - Prob. 10ECh. 11.4 - Prob. 11ECh. 11.5 - Prob. 1ECh. 11.5 - Prob. 2ECh. 11.5 - Prob. 3ECh. 11.5 - Prob. 4ECh. 11.5 - Prob. 5ECh. 11.5 - Prob. 6ECh. 11.5 - Prob. 7ECh. 11.5 - Prob. 8ECh. 11.5 - Prob. 9ECh. 11.5 - Prob. 10ECh. 11.5 - Prob. 11ECh. 11.5 - Prob. 12ECh. 11.5 - Prob. 13ECh. 11.5 - Prob. 14ECh. 11.5 - Prob. 15ECh. 11.5 - Prob. 16ECh. 11.5 - Prob. 17ECh. 11.5 - Prob. 18ECh. 11.5 - Prob. 19ECh. 11.5 - Prob. 20ECh. 11.5 - Prob. 21ECh. 11.5 - Prob. 22ECh. 11.5 - Prob. 23ECh. 11.5 - Prob. 24ECh. 11.5 - Prob. 25ECh. 11.5 - Prob. 26ECh. 11.6 - Prob. 1ECh. 11.6 - Prob. 2ECh. 11.6 - Prob. 3ECh. 11.6 - Prob. 4ECh. 11.6 - Prob. 5ECh. 11.6 - Prob. 6ECh. 11.6 - Prob. 7ECh. 11.6 - Prob. 8ECh. 11.6 - Prob. 9ECh. 11.6 - Prob. 10ECh. 11.6 - Prob. 11ECh. 11.6 - Prob. 12ECh. 11.6 - Prob. 13ECh. 11.6 - Prob. 14ECh. 11.6 - Prob. 15ECh. 11.6 - Prob. 16ECh. 11.CR - Prob. 1CRCh. 11.CR - Prob. 2CRCh. 11.CR - Prob. 3CRCh. 11.CR - Prob. 4CRCh. 11.CR - Prob. 5CRCh. 11.CR - Prob. 6CRCh. 11.CR - Prob. 7CRCh. 11.CR - Prob. 8CRCh. 11.CR - Prob. 9CRCh. 11.CR - Prob. 10CR
Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
MATLAB: An Introduction with Applications
Statistics
ISBN:9781119256830
Author:Amos Gilat
Publisher:John Wiley & Sons Inc
Text book image
Probability and Statistics for Engineering and th...
Statistics
ISBN:9781305251809
Author:Jay L. Devore
Publisher:Cengage Learning
Text book image
Statistics for The Behavioral Sciences (MindTap C...
Statistics
ISBN:9781305504912
Author:Frederick J Gravetter, Larry B. Wallnau
Publisher:Cengage Learning
Text book image
Elementary Statistics: Picturing the World (7th E...
Statistics
ISBN:9780134683416
Author:Ron Larson, Betsy Farber
Publisher:PEARSON
Text book image
The Basic Practice of Statistics
Statistics
ISBN:9781319042578
Author:David S. Moore, William I. Notz, Michael A. Fligner
Publisher:W. H. Freeman
Text book image
Introduction to the Practice of Statistics
Statistics
ISBN:9781319013387
Author:David S. Moore, George P. McCabe, Bruce A. Craig
Publisher:W. H. Freeman