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Random variable Y has binomial distribution with n = 20. The sample median seperates the distribution 鈥� 50% are above 25, the probability in binomial distribution should be p = 0.5

The test id upper tailed so the value P-value is,

\(\begin{array}{l}P = P\left( {Y \ge 15} \right) + P(Y \le 5)\\ = 0.021 + 0.021 = 0.042\end{array}\)

Cumulative density function cdf of bionomial random variable X with parameters n and p is

\(\begin{array}{l}B\left( {x;n,p} \right) = P\left( {X \le x} \right) = \sum\limits_{y = 0}^x {b\left( {y;n,p} \right),} \\x = 0,1...,n\end{array}\)

Theorem:

\(b\left( {x;n,p} \right) = \left\{ {\begin{array}{*{20}{c}}{\left( {\begin{array}{*{20}{c}}n\\x\end{array}} \right){p^x}{{\left( {1 - p} \right)}^{ - x}},x = 0,1,2...n}\\{0,otherwise}\end{array}} \right.\)

Hence the value is P=0.042

Depends on significant level, confidence interval will be different. At significant level 0.042, a 95.8% confidence interval is between values 6^th smallest and 6^th largest value

The 6^th smallest value is 14.4 and 6^th largest value if 41.5, so a 95.8 condidence interval is(14.4,41.5)