a)
M = 5043
t = 2.09
sM = √(17482/21) =
381.44
μ = M ± t(sM)
μ = 5043 ± 2.09*381.44
μ = 5043 ± 795.68
95% CI [4247, 5839]
b)
D
normal dist
This Question: 1 pt mimah mansaray&52 10 of 15(9 complete) A random sample of 21 college...
A random sample of 16 college men's basketball games during the last season had an average attendance of 5,062 with a sample standard deviation of 1,757. Complete parts a and b below. a. Construct a 95% confidence interval to estimate the average attendance of a college men's basketball game during the last season.The 95% confidence interval to estimate the average attendance of a college men's basketball game during the last season is from a lower limit of ______ to an...
part b: what assumptions need to be about this population? Arandom sample of 15 college men's basketball games during the last season had an average attendance of 5,085 with a sample standard deviation of 1,781. Complete parts a and below. a. Construct a 90% confidence interval to estimate the average attendance of a college men's basketball game during the last season The 99% confidence interval to estimate the average attendance of a college men's basketball game during the last season...
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