[Q1: Only answer if you ABSOLUTELY know how to solve AND the answers match up ((
ID: 3335793 • Letter: #
Question
[Q1: Only answer if you ABSOLUTELY know how to solve AND the answers match up ((tired of getting wrong replies.) So, please make sure you’re really certain. It’s appreciated.)
Really Explain EACH STEP, including any formulas you used (& how to use the formula), and explain how to compute with a TI-84 preferably/when possible.]
A restaurant wants to test a new in-store marketing scheme in a small number of stores before rolling it out nationwide. The new ad promotes a premium drink that they want to increase the sales of. 9 locations are chosen at random and the number of drinks sold are recorded for 2 months before the new ad campaign and 2 months after. The average difference in the sales quantity (after - before) is -13.099 with a standard deviation of 50.914. When calculating a 95% confidence interval to estimate the true difference in nationwide sales quantity before the ad campaign and after, what is the margin of error?
Question 1 options:
1)
39.136
2)
16.9713
3)
3.2203
4)
31.1104
5)
38.3918
1)
39.136
2)
16.9713
3)
3.2203
4)
31.1104
5)
38.3918
Explanation / Answer
n1 = n2 = n = 9
d-bar = -13.099
s of d-bar = 50.914
% = 95
Degrees of freedom = n - 1 = 8
SE = s/n = 16.97133333
t- score = 2.306004133
Margin of error = t * SE = 39.13596481
Option (1) is the answer.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.