In: Statistics and Probability
A retail company has started a new advertising campaign in order to increase sales. In the past, the mean spending in both the 18–35 and 35+ age groups was at most $70.00
a. Formulate a hypothesis test to determine if the mean spending has statistically increased to more than $70.00.
b. After the new advertising campaign was launched, a marketing study found that the sample mean spending for 400 respondents in the 18–35 age group was $73.65, with a sample standard deviation of $56.60. Is there sufficient evidence to conclude that the advertising strategy significantly increased sales in this age group with significance level of 5%?
c. For 600 respondents in the 35+ age group, the sample mean and sample standard deviation were $73.42 and $45.44, respectively. Is there sufficient evidence to conclude that the advertising strategy significantly increased sales in this age group with significance level of 5%?
please show work on excel!!!
The Excel output is:
The formulas are:
µ | 70 | |
s | 56.6 | |
n | 400 | |
s/√n | 2.83 | |
(b) | x | 73.65 |
t = (x - µ)/s/√n | 1.29 | |
p-value | 0.0989 | |
µ | 70 | |
s | 45.44 | |
n | 600 | |
s/√n | 1.85508 | |
(c) | x | 73.42 |
t = (x - µ)/s/√n | 1.84 | |
p-value | 0.0330 |
µ | 70 | |
s | 56.6 | |
n | 400 | |
s/√n | =C2/SQRT(C3) | |
(b) | x | 73.65 |
t = (x - µ)/s/√n | =(C6-C1)/C4 | |
p-value | =TDIST(C7,399,1) | |
µ | 70 | |
s | 45.44 | |
n | 600 | |
s/√n | =C11/SQRT(C12) | |
(c) | x | 73.42 |
t = (x - µ)/s/√n | =(C15-C10)/C13 | |
p-value | =TDIST(C16,399,1) |