A telemarketing firm has studied the effects of two factors on the response to i

Question

A telemarketing firm has studied the effects of two factors on the response to its television advertisements. The first factor is the time of day at which the ad is run, while the second is the position of the ad within the hour. The data in the following table, which were obtained by using a completely randomized experimental design, give the number of calls placed to an 800 number following a sample broadcast of the advertisement. If we use Excel to analyze these data, we obtain the output in the table below.

(b) Test the significance of time of day effects with = .05.

  F = 1209.02, p-value = less than .05; (Click to select)do not rejectreject H0: time of day is important

(c) Test the significance of position of advertisement effects with = .05.

  F = 149.14, p-value = less than .001; (Click to select)do not rejectreject H0: position of the ads is important

(d) Make pairwise comparisons of the morning, afternoon, and evening times by using Tukey simultaneous 95 percent confidence intervals. (Round your answers to 2 decimal places. Negative amounts should be indicated by a minus sign.)

(e) Make pairwise comparisons of the four ad positions by using Tukey simultaneous 95 percent confidence intervals. (Round your answers to 2 decimal places. Negative amounts should be indicated by a minus sign.)

(f) Which time of day and advertisement position maximizes consumer response? Compute a 95 percent (individual) confidence interval for the mean number of calls placed for this time of day/ad position combination. (Round your answers to 2 decimal places.)

The Telemarketing Data and the Excel Output of a Two-Way ANOVA Position of Advertisement Time of Day On the Hour On the Half-Hour Early in Program Late in Program 10:00 morning 42 36 62 51 37 41 68 47 41 38 64 48 4:00 afternoon 62 57 88 67 60 60 85 60 58 55 81 66 9:00 evening 100 97 127 105 96 96 120 101 103 101 126 107

Explanation / Answer

b. Per rejection rule, based on p value, one can reject null hypothesis if p value for obtained test statistic is less than given significance level, usually alpha=0.05. Here, p value is less than 0.05, therefore, reject H0.

c. The p value corresponding to F=149.14 is less than alpha=0.05, therefore, reject H0.

d. The Tukeysimultaneous 95% c.i are as follows:

muA-muB=(xbarA-xbarB)+-qalpha/sqrt 2.s sqrt[1/nA+1/nB], where, s is sqrt MSE=sqrt 8.917=2.986, xbar is sample mean, qalpha=3.53 (given), and n is sample size for group A denoting morning and group B denoting afternoon.

=(47.92-66.58)+-3.53*2.986 sqrt[1/12+1/12]

=(-22.96, -14.36)

muA-muC=(47.92-106.58)+-3.53*2.986 sqrt[1/12+1/12]

=(-62.96, -54.36)

muB-muC=(66.58-106.58)+-3.53*2.986 sqrt[1/12+1/12]

=(-44.30, -35.70)

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