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Question: Consumers鈥 attitudes toward advertising. The two most common marketing tools used for product advertising are ads on television and ads in a print magazine. Consumers鈥 attitudes toward television and magazine advertising were investigated in the Journal of Advertising (Vol. 42, 2013). In one experiment, each in a sample of 159 college students were asked to rate both the television and the magazine marketing tool on a scale of 1 to 7 points according to whether the tool was a good example of advertising, a typical form of advertising, and a representative form of advertising. Summary statistics for these 鈥渢ypicality鈥 scores are provided in the following table. One objective is to compare the mean ratings of TV and magazine advertisements.

a. The researchers analysed the data using a paired samples t-test. Explain why this is the most valid method of analysis. Give the null and alternative hypotheses for the test.

b. The researchers reported a paired t-value of 6.96 with an associated p-value of .001 and stated that the 鈥渕ean difference between television and magazine advertising was statistically significant.鈥 Explain what this means in the context of the hypothesis test.

c. To assess whether the result is 鈥減ractically significant,鈥 we require a confidence interval for the mean difference. Although this interval was not reported in the article, you can compute it using the information provided in the table. Find a 95% confidence interval for the mean difference and interpret the result. What is your opinion regarding whether the two means are 鈥減ractically significant.鈥

Source: H. S. Jin and R. J. Lutz, 鈥淭he Typicality and Accessibility of Consumer Attitudes Toward Television Advertising: Implications for the Measurement of Attitudes Toward Advertising in General,鈥 Journal of Advertising, Vol. 42, No. 4, 2013 (from Table 1)

Short Answer

Expert verified

a.

b. The null hypothesis is rejected.

c. The 95% confidence interval for the mean difference is (0.322 to 0.578).

Step by step solution

01

(a) Null and alternative hypothesis of the test

The researchers examined parred information since two separate evaluations were acquired from each student, i.e., each student supplied two ratings. As a result, a television mating commercial, as well as magazine scores, are linked as well as reliant.

02

(b) Context of a hypothesis test

The p-value < 0.001 of the discovery is significant. The null hypothesis is rejected. We may infer that the average difference in advertising among magazines and television was statistically significant.

03

(c) Practically significant

Let 105 be

with

The 95% confidence interval for the mean difference is (0.322 to 0.578). We may infer that the two means are statistically significant because the confidence intervals do not include 0.

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Most popular questions from this chapter

Predicting software blights. Relate to the Pledge Software Engineering Repository data on 498 modules of software law written in 鈥淐鈥 language for a NASA spacecraft instrument, saved in the train. (See Exercise 3.132, p. 209). Recall that the software law in each module was estimated for blights; 49 were classified as 鈥渢rue鈥 (i.e., the module has imperfect law), and 449 were classified as 鈥渇alse鈥 (i.e., the module has corrected law). Consider these to be Arbitrary independent samples of software law modules. Experimenters prognosticated the disfigurement status of each module using the simple algorithm, 鈥淚f the number of lines of law in the module exceeds 50, prognosticate the module to have a disfigurement.鈥 The accompanying SPSS printout shows the number of modules in each of the two samples that were prognosticated to have blights (PRED_LOC = 鈥測es鈥) and prognosticated to have no blights (PRED_LOC = 鈥渘o鈥). Now, define the delicacy rate of the algorithm as the proportion of modules. That was rightly prognosticated. Compare the delicacy rate of the algorithm when applied to modules with imperfect law with the delicacy rate of the algorithm when applied to modules with correct law. Use a 99-confidence interval.

DEFECT*PRED_LOC crosstabulation


PRED_LOC
total
noyes

DEFECT False

True

total

440

29

429

49

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69

449

49

498

Independent random samples from normal populations produced the results shown in the next table.

Sample 1


Sample 2

1.23.11.72.83.0

4.22.73.63.9

a. Calculate the pooled estimate of 2.

b. Do the data provide sufficient evidence to indicate that 2&驳迟;渭1? Test using 伪=.10.

c. Find a 90% confidence interval for (12).

d. Which of the two inferential procedures, the test of hypothesis in part b or the confidence interval in part c, provides more information about (12)?

Drug content assessment. Refer to Exercise 8.16 (p. 467)and the Analytical Chemistry (Dec. 15, 2009) study in which scientists used high-performance liquid chromatography to determine the amount of drug in a tablet. Recall that 25 tablets were produced at each of two different, independent sites. The researchers want to determine if the two sites produced drug concentrations with different variances. A Minitab printout of the analysis follows. Locate the test statistic and p-value on the printout. Use these values =.05and to conduct the appropriate test for the researchers.

Test and CI for two Variances: Content vs Site

Method

Null hypothesis 12=1

Alternative hypothesis 121

F method was used. This method is accurate for normal data only.

Statistics

Site N St Dev Variance 95% CI for St Devs

1 25 3.067 9.406 (2.195,4.267)

2 25 3.339 11.147 (2.607,4.645)

Ratio of standard deviation =0.191

Ratio of variances=0.844

95% Confidence Intervals

Method CI for St Dev Ratio CI Variance Ratio

F (0.610, 1.384) (0.372, 1.915)

Tests

Method DF1 DF2 Test statistic p-value

F 24 24 0.84 0.681

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