I) Since sample is drawn randomly and independently, assumption for constructing confidence interval is met. The confidence interval is
mu -+ t* sigma / sqrt(n)
Where mu = sample mean
sigma =s standard deviation
n is sample size and t is 90% t value at n= (20-1)= 19 degree of freedom.
Here, mu = 1.55
Sigma = 1.3168
ii) 90% confidence interval is
1.55 -+ 1.729*(1.32/sqrt(20))
1.55 -+ 0.51
(1.04, 2.06)
iii) Thus we are 90% confident that number of credit card that a future WKU student has is between (1.04,2.06).
2. (12 pts) A random sample of 20 WKU students is collected and the number of credit cards students have is recorded below n 20 0 0 0 0 Construct a 90% confidence interval to predict the number...
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