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A market researcher for an automobile company suspects differences in preferred color between male and female buyers. Advertisements targeted to different groups should take such differences into account, if they exist. The researcher examines the most recent sales information of a particular car that comes in three colors. Use Table 3.



Gender of Automobile Buyer

Color Male Female
Silver 473 296
Black 549 309
Red 496 373
a. Choose the competing hypotheses to determine whether color preference depends on gender.
H0: Color preference is dependent of gender.; HA: Color preference is independent on gender.
H0: Color preference is independent of gender.; HA: Color preference is dependent on gender

b. Find the critical value at the 1.0% significance level. (Round intermediate calculations to 4 decimal places and your final answer to 2 decimal places.)
Critical value


c. Compute the value of the test statistic. (Do not round intermediate calculations. Round your answer to 2 decimal places.)
Test statistic


d-1. What is the conclusion?
Do not reject H0; color preference is dependent on gender
Do not reject H0; color preference is not dependent on gender
Reject H0; color preference is not dependent on gender
Reject H0; color preference is dependent on gender
d-2.
Does your conclusion suggest that the company should target advertisements differently for males versus females?

Yes
No

Respuesta :

fortuneonyemuwa

Answer:

Step-by-step explanation:

a)

[tex]H_0[/tex]: Rows are columns of given table are independent

[tex]H_1[/tex]: Rows and columns of given table are Not independent

(0)            Male            Female             Total

silver         473                296                 769

Black         549                309                 858

Red            496               373                   869

Total         1518                976                   2496

We calculate expected frequencies by [tex]E=\frac{(Row\, Total)(Column\, Total)}{Grand\, Total}[/tex]

(E)                                    col-1                                               col-2

Row-1    [tex]\frac{769+1518}{2496}=467.69[/tex]      [tex]\frac{769+978}{2496}=301.31[/tex]

Row-2    [tex]\frac{858+1518}{2496}=521.81[/tex]      [tex]\frac{858+978}{2496}=336.19[/tex]

Row-3    [tex]\frac{869+1518}{2496}=528.50[/tex]      [tex]\frac{869+978}{2496}=340.50[/tex]

To calculate chi-square test statistic we calculate this formula on each cell:

[tex]\frac{(observed\, count-expected\, count)^2}{expected\, count}[/tex]

observed count(o)          expected count(E)               [tex]\frac{(O-E)^2}{E}[/tex]

      473                                  467.685                                              0.0604

      296                                  301.315                                               0.09375

      549                                  521.812                                                1.416524

      309                                  336.188                                                2.198654

      496                                  528.502                                                1.998867

      373                                  340.498                                                 3.102537

                          [tex]X^2=\sum \frac{(O-E)^2}{E}=8.870733[/tex]

b)

Critical value [tex]X^2_{0.01}=9.210[/tex]

[tex]X^2_{0.01}[/tex] has an area of 0.01 to its right. To find [tex]X^2_{0.01}[/tex], open any Chi-square table, go across from row [tex]df=v=2[/tex] and down from colum 0.01 and get 9.210

c)

Test statistic X^2=8.871

Significance level = 0.01

Degrees of freedom, [tex]df=(r-1)(c-1)=(3-1)(2-1)=2[/tex]

d1)

To get p-value, we use excel function

CHISQ.DIST.RT(8.870733 , 2) = 0.011850723=0.0119

Since test statistic is not greater than critical value, we fail to reject null hypothesis. At 1% level, there is NOT sufficient evidence that rows and columns are not independent, i.e, it is plausible that rows and columns are independent.

Do not reject [tex]H_0[/tex]: color preference is not dependent on gender

d2)

No

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