Suppose the marketing research firm would like to examine if the social networking site that a person primarily uses is influenced by his or her age. In a randomly drawn sample, 369 social network users were asked which site they primarily visited. At the 0.05 level of significance, can we conclude that the two variables are related? These data are presented in the following table along with each person’s age group: Age (Years)FacebookTwitterLinkedIn10-17722018-3444542535-5440384455 and older263831 a. What test should you run? b. Select the correct hypothesis statements. H0: H1: c. Compute the value of the test statistic? (Round your answer to 2 decimal places. Negative values should have a minus sign in front of them)d. Determine the p-value (Round your final answer to 4 decimal places.) e. What is your decision regarding the null hypothesis? multiple choice Fail to reject the null.Reject the null. f. Interpret the result.

The data shows the sample of 369 social network users were asked which site they primarily visited.a. In this study, the test is about to check whether the two variables age and social network are related or not. Thus, the test should be run is chi-square independence test.b. The hypothesis is, Null hypothesis: H0: The two variables age and social network are not related. Alternative hypothesis: H1: The two variables age and social network are related.c. The expected value formula is, E=Row totalColumn totalGrand total The expected values are, Facebook Twitter LinkedIn Total 10-17 9.20 11.95 7.86 29 18-34 39.00 50.67 33.33 123 35-54 38.68 50.25 33.06 122 55 and older 30.12 39.13 25.75 95 Total 117 152 100 369 The value of the test statistic is, χ2=∑Observed-Expected2Expected=7-9.2029.20+22-11.95211.95+0-7.8627.86+44-39239+54-50.67250.67+25-33.33233.33+40-38.68238.68+38-50.25250.25+44-33.06233.06+26-30.12230.12+38-39.13239.13+31-25.75225.75=0.524+8.462+7.859+0.641+0.219+2.083+0.045+2.988+3.618+0.564+0.033+1.073=28.110 Thus, the value of the test statistic is 28.11.d. The degrees of freedom is, df=r-1c-1=4-13-1=6 Computation of p-value: The p-value of χ2-distribution for test statistic 28.110 at 6 degrees of freedom can be obtained using the excel formula “=CHISQ.DIST.RT(28.110, 6)”. The p-value is 0.0001. Thus, the p-value is 0.0001.e. Decision rule: If p-value ≤ α, then reject the null hypothesis. Decision: The P-value (=0.0001) is less than the level of significance (=0.05). Based on decision rule, reject the null hypothesis.f. The null hypothesis is rejected. Thus, it can be concluded that the two variables age and social network are related.

Answered: Suppose the marketing research firm…

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

Suppose the marketing research firm would like to examine if the social networking site that a person primarily uses is influenced by his or her age. In a randomly drawn sample, 369 social network users were asked which site they primarily visited. At the 0.05 level of significance, can we conclude that the two variables are related? These data are presented in the following table along with each person’s age group:

Age (Years)	Facebook	Twitter	LinkedIn
10-17	7	22	0
18-34	44	54	25
35-54	40	38	44
55 and older	26	38	31

a. What test should you run?

b. Select the correct hypothesis statements.

H₀:

H₁:

c. Compute the value of the test statistic? (Round your answer to 2 decimal places. Negative values should have a minus sign in front of them)

d. Determine the p-value (Round your final answer to 4 decimal places.)

e. What is your decision regarding the null hypothesis?

multiple choice