question 1. A marketing specialist wants to estimate the average amount spent by visitors to an online retailer's newly-designed website. From the data in a preliminary study she guesses that the standard deviation of the amount spent is about 16 dollars.
How large a sample should she take to estimate the mean amount spent to within 4 dollars with 95% confidence? (Round your answer up to the next largest integer).
______
question 2. Pepsi wants to use this technique to determine if they should bid for an exclusivity agreement with NCSU. Such an agreement would bind NCSU to sell Pepsi soft drink products exclusively on campus.
To determine if an exclusivity agreement would be profitable, Pepsi must estimate the total number of 12 ounce cans (or their equivalent) that NCSU students would buy per week. NCSU's Office of Registration and Records informs Pepsi that N = 28,000 different students are on campus at least once during a typical week. Pepsi randomly samples 500 NCSU students and asks them to record the number of 12 ounce cans of soda (or their equivalent) that they purchase on campus over the next 7 days. The results are in this Excel file cans of soda per week.
Use the data below to calculate a 99% confidence interval for the total number of 12 ounce cans of soda that NCSU students will purchase on campus in a week.
lower limit of 99% confidence interval
upper limit of 99% confidence interval.
Amount of Students: 500
Mean: 1.316
Sample Standard Deviation: 1.114733
_____ lower limit of 99% confidence interval
_____ upper limit of 99% confidence interval.
question 3. The local school board wanted to estimate the mean size μ of high school AP statistics classes in their school district. The board obtained the following random sample of class sizes:
24 29 29 23 25 24 33 26 24 29 27 26 22 32 24 21 19 29 31 26 29 31 16
Use the data above to construct a 90% confidence interval for the mean class size μ. (calculate y and s to 3 decimal places).
___ lower bound of confidence interval
____ upper bound of confidence interval
Question 1:
= 0.05
From Table, critical values of Z = 1.96
= 16
e = 4
Substituting, we get:
Question 2:
SE = s/
= 1.114733/ = 0.0499
= 0.01
ndf = 500 - 1 = 499
From Table, critical values of t = 2.5857
Lower limt of confidence interval = 1.316 - (2.5857 X 0.0499) = 1.316 - 0.1290 = 1.1870
Upper limit of confidence interval =1.316 + 0.1290 = 1.4450
Question 3:
From the given data the following statistics are calculated:
n = 23
= 599/23 = 26.0435
s = 4.2478
SE = s/
= 4.2478/ = 0.8857
= 0.10
ndf = 23 - 1= 22
From Table, critical values of t = 1.7171
Lower bound of confidence interval = 26.0435 - (1.7171 X 0.8857) = 26.0435 - 1.5208 = 23.0667
Upper bound of confidence interval = 26.0435 + 1.5208 = 27.5643
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