A manager at LLD Records is investigating the company\'s market research techniq

Question

A manager at LLD Records is investigating the company's market research techniques. She learns that much of the market research of college students is done during promotions on college campuses. She also learns that there are other methods of performing market research (for instance, over the phone, in a mall, etc.). In all cases, for each new CD that LLD Records releases, the company solicits an "intent-to-purchase" score from the student, with being the lowest score ("no intent to purchase") and being the highest score ("full intent to purchase"). The manager finds some information on a soon-to-be-released CD. The information details the intent-to-purchase scores from each of several groups of college students, with each group being questioned via a different method. Based on this information, the manager is able to perform a one-way, independent-samples ANOVA test of the hypothesis that the mean intent-to-purchase score for this CD is the same no matter the method of score collection. This test is summarized in the ANOVA table below. Fill in the missing entry in the ANOVA table (round your answer to at least two decimal places), and then answer the questions below. Source of VariationDegrees of FreedomSum of SquaresMean SquareF statisticTreatments (Between Groups)2524.64262.32 Error (Within Groups)695674.182.23Total716198.74How many intent-to-purchase scores were examined in all? For the ANOVA test, it is assumed that the population variances of intent-to-purchase scores are the same no matter the method of score collection. What is an unbiased estimate of this common population variance based on the sample variances? What is the p-value corresponding to the F statistic for the ANOVA test? Round your answer to at least three decimal places. Based on this ANOVA, can we conclude that there are differences in the mean intent-to-purchase scores (for this CD) among the different methods of collection? Use the 0.01 level of significance. YesNo ClearUndoHelp

Explanation / Answer

Solution-

ANOVA Table is-

Variance component is given by-

S2 total = S2t + S2e

= ( MStreatment - MSerror) / 3

= 60.029

and S2e = 82.2333

then S2 total = 60.029 + 82.233

= 142.262

Test Statistic is calculated mean square factor / mean square error = 262.32/82.233

= 3.1899

P-value will be P[ F( 2, 69 ) > 3.1899 ] = 0.0473

As p-value is less than .05 then we can say that null hypothesis is rejected at 5% level of significance and we can conclude that there are differences in the mean intent-to-purchase scores (for this CD) among the different methods of collection.

Answer

TY!

Source of variation sum of squares degree of freedom mean square Factor( between ) 524.64 2 262.32 Error (within) 5674.1 69 82.2333 Total 71

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.