Question

“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. (You may find it useful to reference the appropriate table: z table or t table)

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

Click here for the Excel Data File

(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: μ₁ − μ₂ = 0; H_A: μ₁ − μ₂ ≠ 0

• H₀: μ₁ − μ₂ ≥ 0; H_A: μ₁ − μ₂ < 0

• H₀: μ₁ − μ₂ ≤ 0; H_A: μ₁ − μ₂ > 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?

Answer 1

The statistical software output for this problem is :

(a) Option A is correct.

b - 1) Test statistics = -7.249

b-2) p-value < 0.01

c)Reject the null hypothesis