Question

In: Math

Researchers for an advertising company are interested in determining if people are more likely to spend...

Researchers for an advertising company are interested in determining if people are more likely to spend more money on beer if advertisers put more beer ads on billboards in a neighborhood in Philadelphia. They estimate that 250 people will view their billboard in one week. They determine that the total number of residents in the neighborhood is 600. So, the residents who have not viewed the billboard are in the control group. The researchers determine that those who did view the ads spent $24 per week on beer and those who did not view the ads spent $16 per week on beer.

Calculate the treatment effect.

Calculate the standard error (hint: the researchers determined the Std. Dev. for the control is $1.3 and the Std. Dev. for the treatment is $0.90)

Calculate the t-statistic.

Can we reject OR fail to reject the null hypothesis? Explain why.

Expert Solution

Given -

For Control -

N = 350

mean = $16

standard deviation = $1.3

For Test -

N = 250

mean = $24

standard deviation = $0.9

A -

treatment effect -

increase in sales/original sales

= 24 - 16/16

= 50%

=> There is 50% lift in sales or the treatment effect is 50%

B -

Since the population standard deviation is seldom known, the standard error of the mean is usually estimated as the sample standard deviation divided by the square root of the sample size.

standard error of control = 1.3/sqrt(350)

= $0.069

standard error of test = 0.9/sqrt(250)

= $0.056

standard error of difference = sqrt(1.3²/350 + 0.9²/250)

= 0.0898

C -

t -test = (sample mean 1 – sample mean 2)/[ sqrt ( s1^2/n1 + s2^2/n2) ]

= 24 - 16/0.0898

= 8/0.0898

= 89.08

D -

This is a one tailed hypothesis since we know we have to calculate whether increase in sales is significant or not

Null hypothesis - There is no significant increase in sales from treatment group as compared to the control group

Alternate hypothesis - There is significant increase in sales from treatment group as compared to the control group

The t critical value at 0.001 significance level and degrees of freedom = 598 is 2.6

The result is significant even at 0.001 significance level

Hence, we can reject the null hypothesis and go with our alternate hypothesis

milcah answered 6 months ago

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