# Question: Example 4 on whether dogs can detect bladder cancer by

Example 4, on whether dogs can detect bladder cancer by selecting the correct urine specimen (out of seven), used the normal sampling distribution to find the P-value. The normal distribution

P-value approximates a P-value using the binomial distribution. That binomial P-value is more appropriate when either expected count is less than 15. In Example 4, n was 54, and 22 of the 54 selections were correct.

a. If H0: p = 1/7 is true, X = number of correct selections has the binomial distribution with n = 54 and p = 1/7. Why?

b. For Ha: p 7 1/7, with x = 22, the P-value using the binomial is P(22) + P(23) + g + P(54), where P (x) denotes the binomial probability of outcome x with p = 1/7. (This equals 0.0000019.) Why would the P-value be this sum rather than just P(22)?

P-value approximates a P-value using the binomial distribution. That binomial P-value is more appropriate when either expected count is less than 15. In Example 4, n was 54, and 22 of the 54 selections were correct.

a. If H0: p = 1/7 is true, X = number of correct selections has the binomial distribution with n = 54 and p = 1/7. Why?

b. For Ha: p 7 1/7, with x = 22, the P-value using the binomial is P(22) + P(23) + g + P(54), where P (x) denotes the binomial probability of outcome x with p = 1/7. (This equals 0.0000019.) Why would the P-value be this sum rather than just P(22)?

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