QUESTION

# Critical Thinking 6

Option 1: Springdale Shopping Survey

Instructions

The major shopping areas in the community of Springdale include Springdale Mall, West Mall, and the downtown area on Main Street. A telephone survey has been conducted to identify strengths and weaknesses of these areas and to find out how they fit into the shopping activities of local residents. The 150 respondents were also asked to provide information about themselves and their shopping habits. The data are provided in the file SHOPPING. The variables in the survey can be found in the file CODING.

In this exercise, some of the estimation techniques presented in the module will be applied to the Springfield Shopping survey results. You may assume that these respondents represent a simple random sample of all potential respondents within the community, and that the population is large enough that application of the finite population correction would not make an appreciable difference in the results.

Managers associated with shopping areas like these find it useful to have point estimates regarding variables describing the characteristics and behaviors of their customers. In addition, it is helpful for them to have some idea as to the likely accuracy of these estimates. Therein lies the benefit of the techniques presented in this module and applied here.

1. Item C in the description of the data collection instrument lists variables 7, 8, and 9, which represent the respondent’s general attitude toward each of the three shopping areas. Each of these variables has numerically equal distances between the possible responses, and for purposes of analysis they may be considered to be of the interval scale of measurement. Determine the point estimate, then construct the 95% confidence interval for μ7= the average attitude toward Springdale Mall. Repeat part (a) for μ8 and μ9, the average attitudes toward Downtown and West Mall, respectively. Given the breakdown of responses for variable 26 (sex of respondent), determine the point estimate, construct the 95% confidence interval for π26= the population proportion of males. Given the breakdown of responses for variable 28 (marital status of respondent), determine the point estimate, and then construct the 95% confidence interval for π28 = the population proportion in the “single or other” category. Assume the managers have requested estimates of the mean attitudes towards each mall with a margin of error of 0.05 for each mall. If the managers want to have 95% confidence that the sample mean will fall within this margin of error, how large should the sample size be for each mall?

Paper Requirements

Write a report that uses the Written Assignment Requirements under the heading Expectations for CSU-Global Written Assignments found in the CSU-Global Guide to Writing and APA Requirements. Items that should be included, at a minimum, are a title page, an introduction, a body that answers the questions posed in the problem, and a conclusion paragraph that addresses your findings and what you have determined from the data and your analysis. As with all written assignments, you should have in-text citations and a reference page too. Please include any tables of calculations, calculated values and graphs associated with this problem in the body of your assignment response.

NOTE: You MUST submit your Excel file with your report. This will aid in grading with partial credit if errors are found in the report.

• @
• 7 orders completed

\$25.00

*** major ******** ***** ** *** ********* of ********** ******* Springdale **** West **** *** *** ******** area ** **** Street * telephone ****** has been ********* ** ********

or
• @
• 1 order completed

\$30.00

* * **** * ** *** *********** OF *** DATA COLLECTION INSTRUMENT ***** ********* * * AND * ***** REPRESENT THE ****************** ******* ATTITUDE ****** **** OF *** ***** ******** ***** **** ** ***** VARIABLES *** *********** ***** DISTANCES ******* *** ******** ********* *** *** PURPOSES ** ******** **** *** ** ********** ** BE ** *** INTERVAL ***** ** *********** *

*********

*** ***** ******** ******* ******** ****** Springdale ***** Let ******************

**

be *********** and *********** *********** ***** ****** ********* **** * *** f ** ******** ********* ***** ******** ******** are *** unknown population parameters (characteristics of ********* For ******* * ****** *** has ****************** ***** and σ2 (the ********** Parameters ** interest

are

********

*

μ *** ******** * ******** that is ******* * (μ ********* ** *** ********** of **** *********** **** **** ********** ** *** *** (1/n

*******

** ∑ (X-X̅)2 ************ then *** corresponding ********* **** *** * ********** ** *** s2= (1/n

*******

** ***** ********* ********* *** sample mean ******** **** is ***

*

(1/n)∑ xi where ***** ****** *** n=150 ********* ******* ******** ****** ******** *** **** ******* **

*****

******** Where

*****

*** ∑ (x-x̅)2=219 Therefore ******* * ********* *** ***** ******** ******* ******** toward ********

*****

********* ***

sample

**** ******** **** is x̅ * ********** xi ***** *****

******

*** ****** ********* ******* ******** ****** variance *** **** ******* ** ∑ ******** ***** n=150 ***

*****

************** *********

s2=114

* Determine *** point estimate ******* ******** ****** Downtown ***** ********* the sample mean

********

**** **

= ********** xi ***** ∑ ****** *** ****** ********* x̅=32

********

Sample ******** *** **** ******* ** ∑ ******** Where n=150 *** ***** ************* Therefore ******* *******

we

have a

random

****** ** ** from ****** ********* ********* * 95% ********** ******** for ***** *********

***

such ******

is

** find *** value ** * **** **** * ********* **** * **** ** = **** From ***

******

******

*

(−zα/2 ≤ Z **** *********** * 095 ***** zα/2 ********** *** ***** of * **** tail **** α/2 This *******

*

= ********** * ***

Hence

* (−196 **** Z **** 196) * ****

**

using *** definition of Z *** solving for **** we ******* *** ******* 196/ √n **** **** ≤ *

*

*************

*

**** *********** * Construct * *** ********** ********

***

*******

****

* ***** ******** ** the **** mean **** ** *** ****** ****

x=7

and

******

******* ***** ** ***** ****** we can *** the ******* ***** *******

***

use

***********

normality

***

*** ************ ** X with **** **** ***

***

approximate ******** error ********************* **** * *** ********** ******** *** 7 ± ***** ***** ********** 02352 ********* ** the ******** ***** 7235) ***** *********** Upper *********** * ********* * *** ********** ******** *** ******* **** * ***** estimate ** the **** mean μ ** *** sample

mean

x=8 and s2=147 ******* n=150 is

*****

****** ** can use the ******* Limit ******* *** *** *********** ********* ***

the

************ **

*

**** ****

****

and *** approximate ******** error ********************* **** a

***

********** ******** *** * ******** ***** (012) ********** ***** resulting ** *** ******** ***** ***** ***** *********** ***** limit=8235 * ********* a *** ********** ******** *** μ=9 Here a ***** ******** ** *** true **** **** ** the sample **** *** *** ****** ******* ***** is

*****

enough ** *** *** *** *******

*****

Theorem and *** *********** ********* *** the ************ of X with **** μ

***

*** ***********

********

***** (147/√150)=012

Thus

a *** ********** ******** *** * ******** (196)

*****

=9± ***** ********* ** *** ******** (8765 9235) Lower *********** ***** *********** ********* ********* * 95% confidence interval for ******* **** a point estimate ** the **** **** **** ** the sample **** *** and ****** ******* ***** is ***** enough we *** use *** *******

Limit

******* *** use *********** ********* ***

***

distribution ** X **** **** **** *** *** *********** ******** error ********************* ****

*

95% confidence

interval

*** *

********

(196)

*****

=7±

*****

resulting ** the ******** ****** ****** Lower

************

Upper

************

*********

* *** ********** ******** for μ=8 **** a point estimate of *** **** mean **** ** the ****** mean *** *** ****** ******* ***** ** large ****** ** *** *** *** ******* Limit ******* *** *** *********** ********* *** *** distribution ** * **** **** ****

and

*** approximate ******** ***** ********************* ****

*

*** ********** ******** *** * ± ***** (009) ********** 01824 ********* ** the

********

****** ******

*****

limit=78176 *****

************

*********

* *** confidence ******** for ******* ****

*

*****

estimate ** *** **** **** **** ** *** ****** mean x=9 and s2=114 ******* n=150 is ***** ****** ** *** use *** ******* ***** ******* *** *** *********** normality for the distribution ** X **** **** **** *** *** *********** ******** ***** ********************* **** * 95%

confidence

******** *** 9 ± ***** (009)

**********

***** ********* in *** ******** (88176 ****** ***** limit=88176 ***** ************ WEST Construct

a

*** **********

********

*** *******

****

*

***** ******** of *** **** mean ****

is

the

sample mean x=7 and ****** ******* ***** ** large ****** ** can *** *** ******* ***** ******* and *** *********** normality *** *** distribution of * with **** **** *** *** *********** ******** ***** (134/√150)=0109 **** a 95% confidence ******** *** * ******** (196) ****** **********

*****

********* ** the ******** ****** ******

Lower

limit=67856 ***** ************ ********* * 95% ********** ******** *** ******* **** a *****

********

** ***

true

**** ****

**

***

******

****

x=8

*** s2=134 ******* n=150 is ***** enough

we

can

***

***

******* ***** ******* *** use *********** ********* for the ************ ** * with **** μ and *** approximate standard ***** (134/√150)=0109 Thus * *** ********** ******** *** * ± (196) ****** ********** ***** ********* ** *** interval ****** ****** ***** ************ ***** ************ ********* * ***

**********

******** for ******* **** a point

estimate

** the **** mean **** ** *** ****** **** x=9 and ****** *******

n=150

** *****

******

we ***

***

***

Central

Limit

******* *** *** *********** ********* *** ***

************

of

* with **** **** *** the *********** ******** error ********************** **** * *** ********** ******** *** * ******** ***** ****** =9± 02144 resulting ** *** ******** ****** ****** ***** ************ ***** limit=92144 * ***** THE ********* ** ********* FOR VARIABLE 26 **** ** *********** DETERMINE ***

*****

******** ********* *** 95% ********** ********

***

******* *** POPULATION ********** OF MALES ******** ********* *** ****** **** estimate ****

is

*** =

**********

** where

*****

******

and ****** Therefore ******* Estimate ****** ********

s2=

****

− ** ∑ ******** Where ***** *** ∑ ************* ********* ******* *** *** ********** ******** for ******* *** population ********** ** males *** ***** **** is to calculate mean *** standard deviation ** *** ****** ** ******* as the mean ****** *** ******** ********* ****** ***

**

******* ** ******* and for *******

*

005 **** the ******** ***** * ***** ***** * 95% ********** ******** for

****

*** ***

*******

******************************* ***

+

******************************** * 16 − ******************** ** + ********************* = (1499 ****** **** *** ** *********** ** **** ** are 95% confident **** *** **** **** μ **** be

*******

**** ***

*****

3 GIVEN THE ********* ** ********* *** VARIABLE ** ******** ****** ** *********** ********* *** *****

ESTIMATE

*** ****

CONSTRUCT

THE *** CONFIDENCE INTERVAL *** ******* THE POPULATION PROPORTION ** *** ************* ** OTHER”

CATEGORY

******** *********

***

****** mean estimate **** is *** = (1/n)∑ xi ***** *****

xi=231

*** ****** ********* ******** ******** ****** variance *** **** − ** ∑ ******** ***** ***** and ***** ************** ********* ******* *** *** ********** ******** for ******* the population ********** ** *** “single ** ************ ********* ***

*****

step is ** calculate **** and ******** ********* ** *** ******

**

******* ** the **** x̅=16 ***

********

deviation ****** *** ** ******* of *******

***

*** ******* *

***

**** *** t-table ***** = ***** ***** * *** ********** ******** *** μ *** x̅ − ******************************* *** * ******************************** *

**

******* 20518(025/√28) ** * 20518(025/√28) * ***** ****** This *** be *********** ** that ** *** 95% confident **** *** **** **** **** **** be ******* **** *** 1697 * ASSUME *** MANAGERS **** *********

*********

** ***

****

ATTITUDES ******* **** **** WITH * ****** ** ***** ** 005 *** **** **** ** ***

********

**** **

HAVE

*** ********** **** *** ****** **** **** **** ****** **** MARGIN ** ERROR ***

*****

****** ***

******

**** BE *** EACH ****** ********* **** ********** level (α) * *** *** **** **********

*

196 **** margin error *** * **** **** ** information ** * we *** *** ****** **** ********* * * (z 2/2) ** 2 * * **** *** 4(005)2 * ******* ***** how ***** ****** the

sample

**** ** *** **** **** should ** ****** ***

or
• @
• 20 orders completed

\$30.00

******

****

*** attached ** the ******** ** the ********

or
• @
• 1164 orders completed

\$30.00

**** particular ********** ** ******* to *** *********** ******** ** a ******** **** located ** *** **** ** Springdale *** ** ** ***** as * ********** Mall ******* in *** ******** **** There are ** *********** have **** ******** *** this analysis *** ***** *** 6 ********* ********* that needed ** ** ********

or
• @
• 16 orders completed