Question

In: Statistics and Probability

10-7. “The See Me” marketing agency wants to determine if time of day for a television...

10-7. “The See Me” marketing agency wants to determine if time of day for a television advertisement influences website searches for a product. They have extracted the number of website searches occurring during a one-hour period after an advertisement was aired for a random sample of 30 day and 30 evening advertisements. The data is shown in the accompanying table and contained in the accompanying Excel file.

Day Searches Evening Searches
96670 118379
97855 111005
96368 100482
98465 122160
98550 117158
101623 101556
95753 98875
102036 104384
99475 110932
103780 101963
97608 123513
99859 102195
101764 111388
97287 116287
98066 119660
95390 112553
96125 98245
96767 98062
99494 101657
102498 106451
99260 104247
102020 105507
102468 118339
96543 109847
102491 123996
96557 122545
102627 113248
95048 104941
96969 111829
95103 114721

(Samples 1 and 2 represent website searches for daytime and nighttime advertisements, respectively.)

a. Set up the hypotheses to test whether the mean number of website searches differs between the day and evening advertisements.

  • H0: μ1μ2 = 0; HA: μ1μ2 ≠ 0

  • H0: μ1μ2 ≥ 0; HA: μ1μ2 < 0

  • H0: μ1μ2 ≤ 0; HA: μ1μ2 > 0



b-1. Calculate the value of the test statistic. Assume that the population variances are unknown but equal. (Negative value should be indicated by a minus sign. Do not round intermediate calculations. Round your answer to 3 decimal places.)



b-2. Find the p-value.
  • p-value ≤ 0.01

  • 0.01 < p-value ≤ 0.02

  • 0.02 < p-value ≤ 0.05

  • 0.05 < p-value ≤ 0.10

  • p-value > 0.10


c. At the 5% significance level, what is the conclusion?

(Reject or Do Not Reject) Ho at the 5% significance level, we conclude that the mean number of website searches (Differs, not differ) between the day and evening advertisements.

Solutions

Expert Solution

Sol:

Instal analysis toolpak in excel

Go to data>data analysis>t test assuming unequal variance

we get

From output:

ANSWER(A)

H0: μ1μ2 = 0; HA: μ1μ2 ≠ 0

ANSWER(b-1)

t=-7.249

ANSWER(B-2)

p=0.0000

p-value ≤ 0.01

ANSWER(C)

Reject Ho at the 5% significance level, we conclude that the mean number of website searches (Differs) between the day and evening advertisements.


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