This year, the company you work for decided to run an advertisement during the S

Question

This year, the company you work for decided to run an advertisement during the Super Bowl for a new product. You were asked to conduct a test to see if sales increased after the Super Bowl. You took a sample of seven days' sales revenue prior to the Super Bowl and another sample of seven days' sales revenue the week following the Super Bowl. You found revenue of $16.1 million with a sample standard deviation of $1.2 million the week before, and r revenue of S20.5 million with a sample standard deviation of $2.3 million the week following the Super Bowl. Using .05 as your level of significance can you conclude that the mean weekly sales revenue increased after the Super BowP Beings the product has not changed, we will assume the population variances are equal, yet unknown. Leave all answer in decimal form.

Explanation / Answer

2.

Formulating the null and alternative hypotheses,              
              
Ho:   u1 - u2   >=   0  
Ha:   u1 - u2   <   0  
At level of significance =    0.05          
As we can see, this is a    left   tailed test.      
Calculating the means of each group,              
              
X1 =    16.1          
X2 =    20.5          
              
Calculating the standard deviations of each group,              
              
s1 =    1.2          
s2 =    2.3          
              
Thus, the pooled standard deviation is given by              
              
S = sqrt[((n1 - 1)s1^2 + (n2 - 1)(s2^2))/(n1 + n2 - 2)]               
              
As n1 =    7   , n2 =    7  
              
Then              
              
S =    1.834393633          
              
Thus, the standard error of the difference is              
              
Sd = S sqrt (1/n1 + 1/n2) =    0.980524641          
              
As ud = the hypothesized difference between means =    0   , then      
              
t = [X1 - X2 - ud]/Sd =    -4.487393602          
              
Getting the critical value using table/technology,              
df = n1 + n2 - 2 =    12          

tcrit =    -   1.782287556      
              
Getting the p value using technology,              
              
p =    0.000371393          
              
As |t| > 1.7822, and P < 0.05, we   REJECT THE NULL HYPOTHESIS.          

Hence, there is significant evidence that the mean weekly sales increased after the super bowl at 0.05 level. [CONCLUSION]

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Hi! If you use another method/formula in calculating the degrees of freedom in this t-test, please resubmit this question together with the formula/method you use in determining the degrees of freedom. That way we can continue helping you! Thanks!

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