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

“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. A portion of the data is shown in the accompanying 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        
a.

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

a) H0: μ1 − μ2 = 0; HA: μ1 − μ2 ≠ 0

b) H0: μ1 − μ2 ≥ 0; HA: μ1 − μ2 < 0

c) H0: μ1 − μ2 ≤ 0; HA: μ1 − μ2 > 0
b-1.

Find the value of the test statistic. Assume the population variances are unknown but equal. (Negative value should be indicated by a minus sign. Round your answer to 3 decimal places.)
  Test statistic =   
b-2.

Assume the population variances are unknown but equal. What are the critical value(s) and the rejection rule? (Negative value should be indicated by a minus sign. Round your answers to 4 decimal places.)
  Reject H0 if tdf> ? or tdf< ? .
c.

At the 5% significance level, what is the conclusion?
  (Click to select)   Reject or Do not reject   H0.At the 5% significance level, we conclude the mean number of website searches   (Click to select)   differs or not differ  between the day and evening advertisements.
Math   Statistics-And-Probability   
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Homework Answers

Answer #1

a)

a) H0: μ1 − μ2 = 0; HA: μ1 − μ2 ≠ 0

b)

value of the test statistic t =-7.249

b-2)

Reject H0 if t >2.002 to t<-2.002

c)

Reject H0.At the 5% significance level, we conclude the mean number of website searches    differs between the day and evening advertisements.

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