An advertising firm has decided to ask 115 customersat each of four local shopping malls if they are willing totake part in a market research survey. At . = 0.05 is thereevidence to show that the proportions are not all thesame.ABDWilling62676970Not Willing53484645What are the expected numbers? (round to 3 decimalplaces)DWillingNot WillingThe hypotheses areHo : PA = pPB = Pc = PpHA: At least one of the proportions is different. (claim)Since a = 0.05 the critical value is 7.815The test value is:(round to 3 decimal places)The p-value is(round to 3 decimal places)So the decision is todo not reject Họreject H,Thus the final conclusion isThere is enough evidence to reject the claim that atleast one of the proportions is different.There is enough evidence to support the claim thatat least one of the proportions is different.There is not enough evidence to support the claimthat at least one of the proportions is different.OThere is not enough evidence to reject the claim thatat least one of the proportions is different.

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Description: An advertising firm has decided to ask 115 customersat each of four local shopping malls if they are willing totake part in a market research survey. At . = 0.05 is thereevidence to show that the proportions are not all thesame.ABDWilling62676970Not

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MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

An advertising firm has decided to ask 115 customers
at each of four local shopping malls if they are willing to
take part in a market research survey. At . = 0.05 is there
evidence to show that the proportions are not all the
same.
A
B
D
Willing
62
67
69
70
Not Willing
53
48
46
45
What are the expected numbers? (round to 3 decimal
places)
D
Willing
Not Willing
The hypotheses are
Ho : PA = pPB = Pc = Pp
HA: At least one of the proportions is different. (claim)
Since a = 0.05 the critical value is 7.815
The test value is:
(round to 3 decimal places)
The p-value is
(round to 3 decimal places)
So the decision is to
do not reject Họ
reject H,
Thus the final conclusion is
There is enough evidence to reject the claim that at
least one of the proportions is different.
There is enough evidence to support the claim that
at least one of the proportions is different.
There is not enough evidence to support the claim
that at least one of the proportions is different.
OThere is not enough evidence to reject the claim that
at least one of the proportions is different.

Expert Solution

Step 1

Solution:

The table containing the number of customers who are willing to take part in market research survey and the number of customers who are not willing to take part in market research survey at each of 4 local shopping malls is given.

Find the row and column totals for each row and column.

	A	B	C	D	Total
willing	62	67	69	70	268
Not willing	53	48	46	45	192

Total	115	115	115	115	n=460

The formula of expected numbers is

$E_{r c} = \frac{r o w t o t a l * c o l u m n t o t a l}{n}$

$E_{11} = \frac{268 * 115}{460} = 67$

$E_{12} = \frac{268 * 115}{460} = 67$

$E_{13} = \frac{268 * 115}{460} = 67$

$E_{14} = \frac{268 * 115}{460} = 67$

$E_{21} = \frac{192 * 115}{460} = 48$

$E_{22} = \frac{192 * 115}{460} = 48$

$E_{23} = \frac{192 * 115}{460} = 48$

$E_{24} = \frac{192 * 115}{460} = 48$

Thus, the table of expected numbers is as follows:

	A	B	C	D
willing	67	67	67	67
Not willing	48	48	48	48

Step 2

It is Chi-square test of homogeneity of proportion.

Test Statistic formula is as follows:

$\begin{array}{rcl} χ^{2} & = & Σ [\frac{{(O_{r c} - E_{r c})}^{2}}{E_{r c}}] \\ = & \frac{{(62 - 67)}^{2}}{67} + \frac{{(67 - 67)}^{2}}{67} + \frac{{(69 - 67)}^{2}}{67} + \frac{{(70 - 67)}^{2}}{67} + \frac{{(53 - 48)}^{2}}{48} + \frac{{(48 - 48)}^{2}}{48} + \frac{{(46 - 48)}^{2}}{48} + \frac{{(45 - 48)}^{2}}{48} \\ = & 0.373134 + 0 + 0.059701 + 0.134328 + 0.520833 + 0 + 0.083333 + 0.1875 \\ = & 1.358831 \end{array}$

After rounding 1.358831 to 3 decimal places, it becomes 1.359.

Thus, test statistic is 1.359.

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