Find the value of tn–1,α needed to construct a 90% upper or lower confidence bound with sample size 12. Round the answer to three decimal places.
solution;
n = Degrees of freedom = df = n - 1 = 12 - 1 = 11
At 90% upper level the t is ,
= 1 - 90% = 1 - 0.90 = 0.1
t
,df = t0.1,11 = 1.363 ( using student
t table)
Find the value of tn–1,α needed to construct a 90% upper or lower confidence bound with...
and i need help finding the upper bound confidence interval as
well
Construct a confidence interval of the population proportion at the given level of confidence. x = 120, n = 1200, 95% confidence Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). The lower bound of the confidence interval is LI. (Round to three decimal places as needed.)
the lower bound =
and upper bound =
and
upper bound!
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