Question

Using Table 2 (average data of 10 brands x 10 attributes), (a) create a positioning map...

Using Table 2 (average data of 10 brands x 10 attributes), (a) create a positioning map by running SPSS, (b) put the overall mean (mean of all attributes per brand) on each brand, and (c) discuss which position must be appropriate for a new brand into the market.

Table 2. Brand Positioning Mean Scores and Dimension Coordinates
Mean scores (B = Brand and A = Attribute)
BRAND A1 A2 A3 A4 A5 A6 A7 A8 A9 A10
B1 3.87 3.13 3.52 3.78 2.61 4.00 3.04 3.13 2.91 2.43
B2 3.30 2.39 3.13 2.48 2.09 3.26 2.91 2.96 2.74 2.65
B3 2.80 2.20 2.70 2.30 1.60 2.80 2.40 3.10 2.60 2.30
B4 2.86 2.86 3.43 2.86 1.86 2.71 2.29 2.00 1.86 1.86
B5 2.53 1.93 2.27 2.27 1.20 2.40 1.93 2.60 2.20 1.73
B6 3.50 3.14 3.29 3.29 2.79 3.50 3.36 3.43 3.21 2.93
B7 4.24 4.00 3.88 4.18 3.82 4.18 3.94 4.00 4.00 3.94
B8 3.30 3.15 3.40 3.00 2.30 3.35 3.05 3.30 2.85 2.85
B9 2.78 1.79 2.51 2.24 1.51 2.91 2.27 2.50 2.09 1.86
B10 2.40 2.00 2.60 2.33 1.47 2.40 2.40 2.47 2.00 1.93
0 0
Add a comment Improve this question Transcribed image text
Answer #1

Answer (a) :

The position map between the products (B1,B2 .....B10) and Attributes (A1,A2,.....A10).

The map for these parameters along with Products are as below.

4.5 3.5 2.5 1.5 0.5 A10 B1 B2 B3 B5 B6 B7 B8 B9 B10

Answer B :The overall mean (mean of all attributes per brand) are as below

BRAND A1 A2 A3 A4 A5 A6 A7 A8 A9 A10 Average
B1 3.87 3.13 3.52 3.78 2.61 4 3.04 3.13 2.91 2.43 3.24
B2 3.3 2.39 3.13 2.48 2.09 3.26 2.91 2.96 2.74 2.65 2.79
B3 2.8 2.2 2.7 2.3 1.6 2.8 2.4 3.1 2.6 2.3 2.48
B4 2.86 2.86 3.43 2.86 1.86 2.71 2.29 2 1.86 1.86 2.46
B5 2.53 1.93 2.27 2.27 1.2 2.4 1.93 2.6 2.2 1.73 2.11
B6 3.5 3.14 3.29 3.29 2.79 3.5 3.36 3.43 3.21 2.93 3.24
B7 4.24 4 3.88 4.18 3.82 4.18 3.94 4 4 3.94 4.02
B8 3.3 3.15 3.4 3 2.3 3.35 3.05 3.3 2.85 2.85 3.06
B9 2.78 1.79 2.51 2.24 1.51 2.91 2.27 2.5 2.09 1.86 2.25
B10 2.4 2 2.6 2.33 1.47 2.4 2.4 2.47 2 1.93 2.20
BRAND Average
B1 3.24
B2 2.79
B3 2.48
B4 2.46
B5 2.11
B6 3.24
B7 4.02
B8 3.06
B9 2.25
B10 2.20

Average 4.50 4.00 3.50 3.24 3.00 2.50 2.00 1.50 1.00 0.50 0.00 4.02 3.06 2.79 2.48 2.46 2.11 2.25 2.20 B1 B2 B3 B4 B5 Вб B7 B

Answer c : The position must be appropriate for a new brand into the market is any position from the lower range to mid range for ech attributes. Thus for any new brand, if it able to achieve any level from lower point to middle point.

Add a comment
Know the answer?
Add Answer to:
Using Table 2 (average data of 10 brands x 10 attributes), (a) create a positioning map...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • REGRESSION 2 • reg bught cigs faminc male Source SS df MS = = Model Residual...

    REGRESSION 2 • reg bught cigs faminc male Source SS df MS = = Model Residual 20477.12 554134.6 3 6825.70666 1,384 400.386272 Number of obs F(3, 1384) Prob > R-squared Adj R-squared Root MSE 1,388 17.05 0.0000 0.0356 0.0335 20.01 - Total 574611.72 1,387 414.283864 bwght Coef. Std. Err. Piti (95Conf. Intervall cigs famine male _cons -.4610457 0913378 .09687980291453 3.113968 1.076396 115.2277 1.20788 -5.050.000 3.32 0.001 2.89 0.004 95.400.000 -.6402212 .0397062 1.002423 112.8582 - 2818702 .1540535 5.225513 117.5972 f) Conduct...

  • Show work if possible please An experiment has a single factor with five groups and five values in each group. freedom....

    Show work if possible please An experiment has a single factor with five groups and five values in each group. freedom. In determining the total variation, there are: determining the among-group variation, there are 4 degrees freedom. In determining the within-group variation, there are 20 degrees of degrees of freedom. Also, note that SSA 96, SsW 120, SST = 216, MSA 24. MSW 6, and FSTAT 4. Complete parts (a) through (d). Click here to view page 1 of the...

  • 1. Two manufacturing processes are being compared to try to reduce the number of defective products...

    1. Two manufacturing processes are being compared to try to reduce the number of defective products made. During 8 shifts for each process, the following results were observed: Line A Line B n 181 | 187 Based on a 5% significance level, did line B have a larger average than line A? *Use the tables I gave you in the handouts for the critical values *Use the appropriate test statistic value, NOT the p-value method *Use and show the 5...

  • Pls both parts for UPVOTE Using Excel create four histograms, properly labeled, depicting the data from...

    Pls both parts for UPVOTE Using Excel create four histograms, properly labeled, depicting the data from the frequency distributions in #a and #b below. a) Using the variable Age, construct TWO frequency distributions to summarize the Ages of students born in the USA and those born outside the USA. Use 5 classes to construct your frequency distribution. For each frequency distribution, show your calculations for the class width and then list the: Lower limits and upper limits Class boundaries Class...

  • Could you please help me with questions 1a-1b please? ( since i could only find the formula neede...

    Could you please help me with questions 1a-1b please? ( since i could only find the formula needed for 1a, if u aren't sure with 1b u can just do 1a but please dont reply "no enough data given " because i have a lil systemical problem with replying the comments) * the first question was asked to complete the anova table ( table 9.1 in the picture ) by using the formulas ( in the pictures) * I have...

  • The median wage for economics degree holders is determined by the following equation: log( wage) = Be...

    The median wage for economics degree holders is determined by the following equation: log( wage) = Be + B educ + B, exper+ B temure + B.age+ B married + u where educ is the level of education measured in years, exper is the job-market experience in years, tenure is the time spend with the current company in years, age is the age in years and married is a dummy variable indicating if a person is married. 935 reg Iwage...

  • Use table C.3 (or B.3 if you are using the 2nd edition) with an α=0.05, the...

    Use table C.3 (or B.3 if you are using the 2nd edition) with an α=0.05, the degrees of freedom_between, and degrees of freedom_within to determine the F critical value needed to make a decision regarding the null hypothesis. What is the F critical value? (degrees of freedom_between = 2, degree of within = 12) Critical Values for the F Distribution TABLE C.3 ues at a .05 level of significance are given in lightface type values at a.01 level of significance...

  • Gain (V/V) R Setting Totals Averages Sample 1 Sample 2 Sample 3 4 ап 7.8 8.1...

    Gain (V/V) R Setting Totals Averages Sample 1 Sample 2 Sample 3 4 ап 7.8 8.1 7.9 3 5.2 6.0 4.3 = 359.3 i=1 j=1 2 4.4 6.9 3.8 1 2.0 1.7 0.8 This is actual data from one of Joe Tritschler's audio engineering experiments. Use Analysis of Variance (ANOVA) to test the null hypothesis that the treatment means are equal at the a = 0.05 level of significance. Fill in the ANOVA table. Source of Variation Sum of Squares...

  • Gain (V/V) R Setting Totals Averages Sample 1 Sample 2 Sample 3 4 ап 7.8 8.1...

    Gain (V/V) R Setting Totals Averages Sample 1 Sample 2 Sample 3 4 ап 7.8 8.1 7.9 3 5.2 6.0 4.3 = 359.3 i=1 j=1 2 4.4 6.9 3.8 1 2.0 1.7 0.8 This is actual data from one of Joe Tritschler's audio engineering experiments. Use Analysis of Variance (ANOVA) to test the null hypothesis that the treatment means are equal at the a = 0.05 level of significance. Fill in the ANOVA table. Source of Variation Sum of Squares...

  • Suppose 1000 coins are tossed. Use the normal curve approximation to the binomial distribution to find...

    Suppose 1000 coins are tossed. Use the normal curve approximation to the binomial distribution to find the probability of getting the following result. Exactly 495 heads Use the table of areas under the standard normal curve given below. Click here to view page 1. Click here to view page 2. Click here to view page 3. Click here to view page 4. Click here to view page 5. Click here to view page 6. The probability of getting exactly 495...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT