After the new advertising campaign was launched, a marketing study found that th
ID: 448376 • Letter: A
Question
After the new advertising campaign was launched, a marketing study found that the sample mean spending for 300 respondents in the 18–35 age group was $75.86, with a sample standard deviation of $50.90. Is there sufficient evidence to conclude that the advertising strategy significantly increased sales in this age group with significance level of 5%?
one sample test for mean
H0: Population mean -<70
H1: Population mean >70
18-35 group= 300 respondents, sample mean spending =$75.86, sample standard deviation=$50.90, Significance level 5%
Therefore, z=$(75.86-70)/50.90/root300 =1.994
Critical Z value=
How can I get a critical Z value from here? Thanks
Explanation / Answer
Since the significane level is 0.95
The Z value can be calcualted using the normal standard deviation tables or using the formula =NORMSINV(0.95)
in excel. One will get the value of critcal Z =1.645.